DUAL LANGUAGE PROGRAMME
ADDITIONAL
MATHEMATICS
Form
RUKUN NEGARA
Bahawasanya Negara Kita Malaysia
mendukung cita-cita hendak;
Mencapai perpaduan yang lebih erat dalam kalangan
seluruh masyarakatnya;
Memelihara satu cara hidup demokrasi;
Mencipta satu masyarakat yang adil di mana kemakmuran negara
akan dapat dinikmati bersama secara adil dan saksama;
Menjamin satu cara yang liberal terhadap
tradisi-tradisi kebudayaannya yang kaya dan pelbagai corak;
Membina satu masyarakat progresif yang akan menggunakan
sains dan teknologi moden;
MAKA KAMI, rakyat Malaysia,
berikrar akan menumpukan
seluruh tenaga dan usaha kami untuk mencapai cita-cita tersebut
berdasarkan prinsip-prinsip yang berikut:
KEPERCAYAAN KEPADA TUHAN
KESETIAAN KEPADA RAJA DAN NEGARA
KELUHURAN PERLEMBAGAAN
KEDAULATAN UNDANG-UNDANG
KESOPANAN DAN KESUSILAAN
(Sumber: Jabatan Penerangan, Kementerian Komunikasi dan Multimedia Malaysia)
Authors
Translators
Editors
Designer
Illustrator
Dr. Wong Mee Kiong
Zaini bin Musa
Azizah binti Kamar
Saripah binti Ahmad
Nurbaiti binti Ahmad Zaki
Zefry Hanif bin Burham@Borhan
Yew Chian Hauo
Lee Chye Mei
Izyani binti Ibrahim
Nur Marliesa Atiera binti Zakaria
Wan Fathiyyah binti Wan Ismail
Nur Shuhada binti Osman
Ng Peck Foong
Ng Ying Tong
2019
BOOK Series NO: 0174
KPM2019 ISBN 978-967-2375-42-5
First Published 2019
© Ministry of Education Malaysia
All rights reserved. No part of this book may
be reproduced, stored in a retrieval system, or
transmitted in any form or by any means, either
electronic, mechanical, photocopying, recording
or otherwise, without the prior permission of
Director General of Education Malaysia, Ministry
of Education Malaysia. Negotiations are subject to
an estimation of royalty or an honorarium.
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• GeoGebra
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• Everyone who has been directly or indirectly
involved in the successful publication of this
Additional Mathematics Form 4 textbook.
This KSSM Form 4 Additional Mathematics Textbook is written based on Dokumen Standard
Kurikulum dan Pentaksiran (DSKP) for Additional Mathematics Form 4. The purpose of Additional
Mathematics Kurikulum Standard Sekolah Menengah (KSSM) is to form individuals who think
mathematically, creatively and innovatively as well as having a good image.
The contents of this textbook integrate six basis of the KSSM structure, knowledge, skills
and values as well as explicitly instilling 21st Century Skills and Higher Order Thinking Skills
(HOTS). This textbook weaves the diversity of teaching and learning strategies to enable students
to understand the contents in-depth as well as sharpen their thinking to a higher level. Through
the complete usage of this book, students will actively participate in inquiry-based learning that
involves experience, investigation and exploration.
Cross-Curricular Elements (CCE) such as the correct usage of language medium, moral
values, patriotism, scientific and technological literacy, creativity and innovation, entrepreneurship,
information technology and financial education are wholly instilled in the formation of contents
of this textbook. A STEM approach is also applied in this book in preparation for students to face
challenges and be competitive at the global stage.
Special Characteristics in this Book and its Functionalities
Stimulus Pages
• Contain interesting photographs and texts that relate to daily life which stimulate students’
thinking.
• Contain Content Standards in ‘What will be learnt?’, learning aims in ‘The significance of this
chapter’, history or general information about the chapter in ‘Did you know?’ and bilingual Key
Words.
The QR code on the front cover of the book contains
explanation of the book themes, the author’s biography
as well as updated information and facts (if any).
InQuirY 1
In pairs
In groups
Individual
Self Practice 1.1
Intensive Practice 1.1
Activities that involve students individually, in pairs or in
groups which encourage students to be actively involved
in the learning process.
Prepares questions to test the students’ understanding
about learnt concepts.
Contains questions to determine students’ mastery of a
learnt topic.
iii
Mathematics
Application
FLASHBACK
Gives problem solving questions as well as working steps
that cover real life situations.
Shows information that is learnt by students.
Mind Challenge
Provides questions that require students to think
creatively and test students’ performance.
Mathematics Museum
Gives explanation about the developments in the history
of mathematics and contributions of mathematicians.
BRAINSTORMING
Provides activities that require discussion amongst
students.
QUICK COUNT
Explains the ways of using the scientific calculator in
mathematical calculations.
Tech Whizz
Gives exposure to students regarding applications of
technology in mathematical learning.
QR
Smart TIPS
Alternative Method
MATHEMATICS
POCKET
Gives exposure to students using mobile devices by
scanning the QR code to get additional information.
Gives mathematical tips that relate to the topic for
student’s use.
Suggests an alternative method to certain questions.
Gives additional information to students to master the
learnt topic more.
Summary OF Chapter
A whole coverage about the learnt chapter.
WRITE YOUR JOURNAL
Apply learnt concepts in daily lives.
Exploring
Brief activities that relate to the learnt topic.
MATHEMATICS
MASTERY PRACTICE
Covers LOTS and HOTS questions to test students’
understanding.
HOTS question is to stimulate students’ higher order
thinking skills.
21st Century Learning
1.1.2
Uses 21st century learning concepts to increase students’
understanding.
Shows the learning standards for each chapter.
Shows the performance level for each question.
iv
iii
vii
Introduction
Formulae
Chapter 1 Functions
1
2
12
20
30
31
33
1.1 Functions
1.2 Composite Functions
1.3 Inverse Functions
Summary of Chapter
Mastery Practice
Exploring Mathematics
Chapter 2 Quadratic Functions
34
36
45
49
65
66
67
2.1 Quadratic Equations and Inequalities
2.2 Types of Roots of Quadratic Equations
2.3 Quadratic Functions
Summary of Chapter
Mastery Practice
Exploring Mathematics
Chapter 3
Systems of Equations
68
70
79
3.1 Systems of Linear Equations in Three Variables
3.2 Simultaneous Equations involving One Linear
Equation and One Non-Linear Equation
Summary of Chapter
Mastery Practice
Exploring Mathematics
Chapter 4 Indices, Surds and Logarithms
4.1 Laws of Indices
4.2 Laws of Surds
4.3 Laws of Logarithms
4.4 Applications of Indices, Surds and Logarithms
Summary of Chapter
Mastery Practice
Exploring Mathematics
Chapter 5 Progressions
5.1 Arithmetic Progressions
5.2 Geometric Progressions
Summary of Chapter
Mastery Practice
Exploring Mathematics
85
86
87
88
90
96
109
122
123
124
125
126
128
139
150
150
151
v
Chapter 6 Linear Law
6.1 Linear and Non-Linear Relations
6.2 Linear Law and Non-Linear Relations
6.3 Applications of Linear Law
Summary of Chapter
Mastery Practice
Exploring Mathematics
Chapter 7 Coordinate Geometry
7.1 Divisor of a Line Segment
7.2 Parallel Lines and Perpendicular Lines
7.3 Areas of Polygons
7.4 Equations of Loci
Summary of Chapter
Mastery Practice
Exploring Mathematics
Chapter 8
Vectors
8.1 Vectors
8.2 Addition and Subtraction of Vectors
8.3 Vectors in a Cartesian Plane
Summary of Chapter
Mastery Practice
Exploring Mathematics
Chapter 9 Solution of Triangles
9.1 Sine Rule
9.2 Cosine Rule
9.3 Area of a Triangle
9.4 Application of Sine Rule, Cosine Rule and Area of a Triangle
Summary of Chapter
Mastery Practice
Exploring Mathematics
Chapter 10 Index Number
10.1 Index Numbers
10.2 Composite Index
Summary of Chapter
Mastery Practice
Exploring Mathematics
Answers
Glossary
References
Index
vi
152
154
162
166
170
171
173
174
176
184
192
200
206
207
209
210
212
221
227
236
237
239
240
242
251
256
263
266
267
269
270
272
279
284
285
287
288
309
311
312
Chapter 2 Quadratic Functions
x=
–b ± !w
b2 – 4ac
2a
Chapter 4 Indices, Surds and
Logarithms
am × an = am + n
am ÷ an = am – n
(am)n = amn
a × !w
b = !w
ab
!w
a
a ÷ !w
b= w
!w
b
logamn = logam + logan
m
loga = logam – logan
n
logamn = n logam
logcb
logab =
logca
Divisor of a Line Segment
nx1 + mx2 ny1 + my2
= m
+n , m+n
Area of triangle
1
= [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)]
2
Area of quadrilaterals
1
= [(x1y2 + x2y3 + x3y4 + x4y1)
2
– (x2y1 + x3y2 + x4y3 + x1y4)]
(
Chapter 8
!
x2 + y2
∼r  = !w
^ r
∼r = ∼r 
∼
Chapter 5 Progression
Arithmetic Progressions
Tn = a + (n – 1)d
n
Sn = [2a + (n – 1)d]
2
n
Sn = [a + l]
2
Geometric Progressions
Tn = arn – 1
a(1 – rn)
Sn =
, r < 1
1–r
a(rn – 1)
, r > 1
Sn =
r–1
a
, r < 1
S∞ =
1–r
Chapter 7 Coordinate Geometry
)
Vectors
Chapter 9 Solution of Triangles
a
b
c
=
=
sin A sin B sin C
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
Area of triangle
1
1
1
= ab sin C = ac sin B = bc sin A
2
2
2
Heron’s Formula
a+b+c
s(s – a)(s – b)(s – c), s =
= !w
2
Chapter 10 Index Number
Q
I = 1 × 100
Q0
∑I w
I = i i
∑wi
Download the free QR code scanner application from Google Play,
App Store or other platforms to your mobile devices. Scan the QR
Code using the application or visit the website as shown on the left to
download PDF files, GeoGebra and complete answers. Then, save the
downloaded files for offline use.
bit.ly/2nFGX42
vii
CHAPTER
Functions
1
What will be learnt?
Functions
Composite Functions
Inverse Functions
List of
Learning
Standards
bit.ly/329x6lO
wORDS
KEY W
Function notation
Undefined function
Absolute value function
Vertical line test
Arrow diagram
Object
Image
Domain
Codomain
Range
Discrete function
Continuous function
Composite function
Inverse function
Horizontal line test
Tatatanda fungsi
Fungsi tidak tertakrif
Fungsi nilai mutlak
Ujian garis
mencancang
Gambar rajah anak
panah
Objek
Imej
Domain
Kodomain
Julat
Fungsi diskret
Fungsi selanjar
Fungsi gubahan
Fungsi songsang
Ujian garis mengufuk
Proton again makes Malaysians
proud by producing a new model,
that is Proton X70 which gives a
high efficiency in fuel usage. Proton
X70 is powered by a 1.8 litre TGDI
(Turbocharged Gasoline Direct
Injection) engine which makes
this model a powerful and yet fuel
saving efficient vehicle. This car
model is categorised as Energy
Efficient Vehicles (EEV) by the
Road Transport Department (RTD).
Do you know that the formula used
by the engineers in measuring the
efficiency is closely related with
functions? For your information,
efficiency of usage of fuel for
10 litres of petroleum is given as
d km
C =
, where C is the rate of
10(l)
usage of fuel and d is the distance
travelled.
Did you Know?
The subject of function was first introduced by
the French mathematician, Rene ­Descartes,
in the year 1637. According to him, a function
is any variable x where its power is a positive
integer.
However, Leonhard Euler (1707-1783),
a mathematician from Switzerland stated that
a function is any equation or formula involving
variables and constants. His idea regarding
functions is similar to what is ­being studied
these days.
For further information:
bit.ly/2B2y33v
SIGNIFICANCE
OF THIS CH
APTER
A function gives rise to simple and accurate
mathematical model in representing a situation
as well as in solving problems faced in our
surroundings. For example:
The height of an individual, h, is a function
related to his/her thigh bones, f. By
substituting the values of f into the function
h, forensic experts are able to estimate the
height of a corpse based on his/her thigh
bones.
Bank officers use the concept of functions
in calculating the interest incurred in a loan
and hence in the instalments of purchasing a
house, a car, a personal or business loan of
their clients.
Scan this QR code
to watch a video on
Proton X70.
bit.ly/2Rnu0Zh
1
CHAPTER 1
1.1
Functions
There are many quantities which depend on one or more variables in our daily lives. Study and
understand the following situations:
You are working as a
temporary cashier and
is paid RM80 daily. The
total payment received
is determined by the
number of days you
worked.
You are buying durians
from a stall. If the cost
of one kilogram of
durian is RM8, the total
amount you need to pay
depends on the weight
of durians you bought.
In mathematics, such situations are examples of functions. From the above examples of
situations, state the meaning of a function.
Explaining functions by graphical representation and notation
Look at the graph of y = 2x + 1 on the right. The relation
between the value of 1 on the x-axis and the value of 3 on the
y-axis can be written as 1 → 3. This indicates that 1 is the first
element and 3 is the last element.
In this case, we can say that 1 is mapped to 3. Similar
to 3 → 7, 5 → 11 and so on. Every point (x, y) on the line is
corresponding to the mapping of x → y where x on the x-axis is
mapped to y on the y-axis.
The relation from part of the mapping x → y can be
represented by an arrow diagram as shown below.
1
3
3
7
5
11
Set X
y
14
12
10
8
6
4
2
0
(x, y)
(5, 11)
(3, 7)
(1, 3)
2 4 6 8
x
Each element x in
set X is mapped to
one and only one
element y in set Y.
Set Y
Thus, this type of relation is known as function or mapping.
In general:
Function relating set X to set Y is a special relation where
each element x  X is mapped to one and only one element
yY
2
1.1.1
If f is a function from set X = {1, 3, 5} to set Y = {3, 7, 11}
and is defined by f : 1 → 3, f : 3 → 7 and f : 5 → 11, element 1 is
known as the object and element 3 is its image. Similarly, 7 and
11 are the images of 3 and 5 respectively. Any element x in set X
that is mapped to one element y in set Y by y = 2x + 1 is written in
function notation as below:
f:x→y
or
f(x) = y
f : x → 2x + 1 or f(x) = 2x + 1
where x is the object and 2x + 1 is the image
Example 1
6
8
20
22
24
(c)
1
3
7
9
(d)
4
2
8
3
18
9
a
4
b
b
10
c
c
12
Solution
(a) This relation is a function because each object has only
one image even though element 22 has no object.
(b) This relation is not a function because it does not satisfy
the condition of being a function, that is each object has
only one image. Note that 18 has two images, that are
18 → 3 and 18 → 9.
(c) This relation is not a function because it does not satisfy
the condition of being a function, that is each object has
only one image. Note that 9 does not have any image.
(d) This relation is a function because each object has only one
image even though element 10 has no object.
1.1.1
1
3
3
7
5
11
FLASHBACK
a
• f : x → 2x + 1 is read as
“function f maps x to
2x + 1”.
• f(x) = 2x + 1 is read as “2x + 1
is the image of x under the
function f ” or “the function
f of x is equal to 2x + 1”.
•
f
x
2x + 1
Set Y
Set X
is read as
“2x + 1 for 1 is 3” and so on.
Are the following relations a function? Explain.
(a)
(b)
2
5
MATHEMATICS
POCKET
Function is a relation of
one-to-one or a relation of
many-to-one.
classification of animals
Crocodiles
Whales
Frogs
Reptiles
Mammals
Amphibians
Relation of one-to-one
favourite subjects
Ali
Rama
Tan
Physics
Chemistry
Relation of many-to-one
3
CHAPTER 1
Functions
CHAPTER 1
How can we determine whether a graph of a relation is a function? When a graph is given, we
use the vertical line test to determine whether the given graph is a function or otherwise. If
the vertical line cuts only at one point on the graph, then the relation is a function. On the other
hand, if the vertical line does not cut the graph at any point or cuts more than one point, then
the graph is not a function.
y
The graph of h
is a function.
h
y
Vertical
line test
The graph of h
is not a function.
x
0
x
0
Vertical
line test
h
Example 2
Which of the following graph represents a function?
y
(a)(b)(c)
y
0
1
x
y
x
–1 0
x
0
1
Solution
y
(a) This graph is a function because when the
vertical line test is carried out, the line cuts the
graph at only one point, except when x = 1 where
the line does not cut any point on the graph.
0
y
(b) This graph is a function
because when the vertical line
test is carried out, the line cuts
the graph at only one point.
1
–1 0
(c) This graph is not a function
because when the vertical line
test is carried out, the line cuts
the graph at two points.
4
y
0
1
x
Vertical
line test
Vertical
line test
x
Vertical
line test
Mind Challenge
x
How many x-intercepts and
y-intercepts can exist in the
graph of a function?
1.1.1
Observe the graph in Example 2(a). The graph is for the function
x
f(x) =
. From the graph in Diagram 1.1, we obtain that
x–1
–
when x → 1 , that is x approaches 1 from the left, f(x) → –∞,
thus the value of f(x) decreases non-terminating. When x → 1+,
that is x approaches 1 from the right, f(x) → ∞, thus the value
of f(x) increases non-terminating. This implies that the graph
will approach but will never touch the line x = 1. Therefore, this
function is not defined at x = 1.
Next, observe the graph in Example 2(b). The graph
represents the function of absolute value f(x) = |x + 1|. The
expression of the absolute value |x| is the numerical value of x
and is defined by:
y
x if x  0
|x| =
f(x) = |x|
–x if x < 0

Thus, when x = –2, |–2| = –(–2)
=2
and when x = 2, |2| = 2.
x
0
Diagram 1.2
The function which is defined by f(x) = |x| has a V-shaped
graph where the vertex is at (0, 0) as shown in Diagram 1.2.
|x| is read as “the modulus of x”.
y
f(x) =
0
x
x–1
x
1
Diagram 1.1
MATHEMATICS
POCKET
Based on Diagram 1.1:
• x → 1– means x
approaches 1 from the left
side on the graph of
x
f(x) =
, x < 1.
x–1
• x → 1+ means x
approaches 1 from the
right side on the graph of
x
f(x) =
, x > 1.
x–1
Example 3
g
Based on the diagram on the right, write the relation for function g
by using the function notation.
–3
4
2
3
5
Solution
The function notation for the function is g : x → x2 or g(x) = x2.
9
25
Self Practice 1.1
1. State whether each of the following relation is a function. State your reason.
(a)(b)(c)
1
square of
2
2
4
6
X
1.1.1
of
1
2
3
7
Y
–2
2
3
M
p
4
9
q
r
N
A
1
5
8
10
B
5
CHAPTER 1
Functions
x
0
1
x
0
0
x
2
3. By using the function notation, express h in terms of x for each of the following arrow diagrams.
(a)(b)(c)
h(x)
x
x
h(x)
h(x)
x
2
3
5
1
2
1
3
1
5
–5
–4
4
5
4
5
–3
–2
1
4
–27
–8
1
64
Determining the domain and range of a function
InQuirY 1
In groups
21st Century Learning
Aim: To explore the domain and range of a discrete function and a continuous function.
Instruction:
1. Each group is required to choose one of the situations below.
y
Total price (RM)
Situation I
Function y = 18x represents the price of
tickets, in RM, for x tickets
purchased by a family to watch a film.
Draw a graph of the function for the
purchasing of 1 to 5 tickets.
90
72
54
36
18
0
y
Situation II
A packet of popcorn contains 60 calories.
y calories is a function of x packets of
popcorn consumed. Draw a graph of the
function for the purchasing of 1 to 4
packets of popcorn.
Calories consumed
CHAPTER 1
2. Determine whether each of the following graphs is a function by using the vertical line test.
y
y
y
(a)(b)(c)
1 2 3 4 5
Number of tickets
x
240
180
120
60
x
1
2
3 4
Number of packets of popcorn
0
6
1.1.1
1.1.2
Functions
CHAPTER 1
y
Distance from
Hilal’s house (m)
Situation III
Hilal cycled a distance of 100 m from his
friend’s house at the speed of 10 ms–1.
With the same speed, Hilal cycled back to
his friend’s house to take his books that
he left behind. Complete the distancetime graph of Hilal’s journey.
100
0
10
Time (seconds)
x
2. Based on the graph drawn, discuss with the members of your group and answer the
following questions.
(a) Is the graph of the function chosen discrete or continuous? Explain.
(b) Identify the domain and range of the graph of function.
3. Present the findings of your group to the class.
From the results of Inquiry 1, it is noticed that points on the graph of the discrete function are real,
separated and not connected by a straight line or a curve. As for the graphs of continuous function,
the points are connected by a straight line or a curve within the given interval. Thus, Situation
I represents a discrete function whereas Situations II and III represent continuous functions.
In general, the domain of a function is the set of possible values of x which defines a
function, whereas range is the set of values of y that are obtained by substituting all the
possible values of x.
Discrete function
f(x)
Range
Continuous function
f(x)
0
Range
Domain
x
0
x
Domain
Look at the arrow diagram of a discrete function f in Diagram 1.3.
f
2x + 1
x
In this function, the elements in set X are mapped to a corresponding
1
3
element in set Y respectively.
2
5
The elements in set X, are the values of x that can be substituted
3
7
in f which is known as the domain whereas elements of set Y, are
4
9
the possible obtained values of function f which is known as the
11
codomain. The elements in set Y that are mapped from X, are the
X
Y
actual obtained values of function f is known as the range.
Domain
Codomain
Thus, we obtain
Diagram 1.3
Domain = {1, 2, 3, 4}
Codomain = {3, 5, 7, 9, 11}
Range = {3, 5, 7, 9}
Next, consider a continuous function f(x) = 2x + 1 that can take all values of x from 1 to 4. Can
you determine the domain, codomain and the range?
1.1.2
7
CHAPTER 1
Example 4
Determine the domain, codomain and range for each of the following functions f.
(a)(b)(c)
f
f(x)
f(x)
y
x
1
3
4
7
9
Y
p
q
r
s
X
6
4
3
2
2
–2
0
2
x
4
0
1
x
3
Solution
(a) Domain = {p, q, r, s}
(b) Domain = {–2, 0, 1, 2, 3, 4}
Codomain = {1, 3, 4, 7, 9} Codomain = {1, 2, 3, 4, 5}
Range = {1, 3, 7, 9} Range = {1, 2, 3, 4, 5}
(c) Domain of f is 0  x  3.
Codomain of f is 2  f(x)  6.
Range of f is 2  f(x)  6.
Alternative Method
Example 5
Function f is defined as f : x → |2x – 1|. Sketch the graph of f in
the domain –1 < x < 2 and state the corresponding range of f
for the given domain.
Solution
The graph of f(x) = |2x – 1| can be sketched by plotting several
points in the domain –1 < x < 2 as shown in the following table.
x
–1
0
y = f(x) = |2x – 1|
3
1
(x, y)
(–1, 3)
(0, 1)
1
2
0
( 12 , 0)
1
2
1
3
(1, 1)
(2, 3)
f(x)
f(x) = |2x – 1|
3
1
1
–1 0 –
2
1, 0
Vertex –
2 2
x
From the graph, the range of f : x → |2x – 1| is 0  f(x)  3.
8
From Example 5, draw the
graph of y = 2x – 1 in the
domain –1  x  2 first.
The graph below the x-axis
is reflected on the x-axis to
obtain the graph of
f(x) = |2x – 1|.
Tech Whizz
By using the GeoGebra
software, draw the graphs
of y = |x|, y = 2|x|, y = 4|x|
1
and y = |x|.
2
What pattern do you
observe? Can you predict
the graphs of y = 8|x| and
1
y = |x|?
4
1.1.2
Self Practice 1.2
1. Determine the domain, codomain and range of the following functions.
g
(a)(b)(c)
f(x)
y
2
3
6
7
10
Y
x
j
k
l
m
4
2
–2
0
2
4
x
X
y
6
y = f(x)
2
–3
0 1
x
5
2. Sketch the graph of each of the following functions in the domain –2 < x < 4. Hence, state
the corresponding range for the given domain.
(b) f(x) = 4 – 2x
(c) f : x → 2x – 5
(a) f : x → x + 1
Determining the image of a function when the object is given and
vice versa
Consider a fruit juicer. When we put oranges into the juicer, orange juice will
be obtained. It is impossible for us to obtain other than orange juice.
Think of this analogy where a function is the machine with input and
output or an object and its image. Thus, if the object x is given and by
inserting it in a function, the corresponding image f(x) can be determined.
Similarly, if the image, f(x) is given, the corresponding object x can also be
determined.
Example 6
Function f is defined by f : x → 3x + 5 , x ≠ 0. Find
x
(a) f(5),
(b) the image of 1 under f,
3
(c) the possible values of x when their image is 8.
Solution
(a) f(5) = 3(5) + 5
5
= 15 + 1
= 16
(b) Given f(x) = 3x + 5 .
x
1
The image of ,
3
f 1 =3 1 + 5
3
3
1
3
= 1 + 15
= 16
() ()
1.1.2
1.1.3
()
Mind Challenge
5
f : x → 3x + , x ≠ 0.
x
Why is x ≠ 0?
2
, x ≠ k,
If f(x) =
x+3
what is the value of k?
(c)
f(x) = 8
Multiply both sides
3x + 5 = 8
of the equation
x
2
with x.
3x + 5 = 8x
2
3x – 8x + 5 = 0
(3x – 5)(x – 1) = 0
x = 5 or x = 1
3
Thus, the possible values of x are
x = 5 and x = 1.
3
9
CHAPTER 1
Functions
CHAPTER 1
Example 7
The diagram on the right shows parts of the graph f (x) = |2x – 3|,
find
(a) the values of f (–2) and f (4),
(b) the values of x such that f (x) = 5,
(c) the values of x which maps to itself,
(d) the domain of f (x) , 1,
(e) the domain of f (x) > 3.
y
f(x) = |2x – 3|
3
Solution
(a) f (–2) = |2(–2) – 3|
(b) f (x) = 5
= |–7|
|2x – 3| = 5
= 7
2x – 3 = –5 or 2x – 3 = 5
f (4) = |2(4) – 3| 2x = –2
2x = 8
= |5| x = –1
x=4
=5
(c) f (x) = x
(d) f (x) < 1
|2x – 3| = x |2x – 3| < 1
2x – 3 = –x or
2x – 3 = x
–1 < 2x – 3 < 1
3x = 3
x = 3
2 < 2x < 4
x = 1
1<x<2
(e) f (x)  3
2x – 3  3
2x – 3  –3 or
2x  0
2x  6
x  0
x 3
x
3–
2
0
QR
The solution of equality
and inequality involving
absolute values.
bit.ly/2Oz1EcZ
Self Practice 1.3
6
1. Function g is defined by g : x → 3 +
, x ≠ 1.
x–1
1
(a) Find the images of –5, –2 and .
2
(b) Given the image of b is 2b, find the possible values of b.
kx – 3
2. Function h is defined by h : x →
, x ≠ 1. Find the value of k such that
x–1
(a) h(2) = 5
(b) h(3) = k
(c) h(k) = k
3. Function f is defined by f : x → |4x – 3|, calculate
1
(a) f (–2) and f 1– 2,
(b) the values of x such that f (x) = 1,
2
(c) the domain of f (x) , 1,
(d) the domain of f (x) . 5.
4. Given g(x) = |6 – 2x|, find the values of x if g(x) = x.
5. Function f is defined by f : x → mx + c. Given f (2) = 7 and f (4) = –1, find
(a) the value of m and of c,
(b) the image of 2 under f,
(c) the value of x that is unchanged under
the mapping of f.
10
1.1.3
Intensive Practice 1.1
Scan the QR code or visit bit.ly/33iJznC for the quiz
1. Which of the following relations are functions? State your reasons.
(a)(b)(c)
4
–4
–1
3
7
a
b
c
d
X
Y
1
5
2
8
X
Y
CHAPTER 1
Functions
2
6
10
9
Y
X
2. By using the vertical line test, determine whether the following graphs are functions.
(a)(b)(c)
y
y
y
0
x
0
x
3. The diagram on the right shows the images for certain elements
of set A.
(a) Is the relation a function? If so, state your reason.
(b) State the domain and range for that relation.
(c) Using the function notation, write one relation between set
A and set B.
4. The diagram on the right shows the graph of the function
f(x) = |2x – 4| the domain 0  x  5. Find
(a) the value of t,
(b) the range of f based on the given domain,
(c) the range of values of x such that f(x)  4.
0
x
2
7
6
–6
–7
49
36
Set B
Set A
y
f(x) = |2x – 4|
t
4
0
2
x
5
5. A stone fell from a height of 81 metres above the ground. The
height of the stone, H metres, after t seconds, is assumed to be
H(t) = 81 – 9t2.
(a) State the height of the stone when
(i) t = 1 second,
3
(ii) t = 1 second,
(iii) t = 2 seconds.
(b) When will the stone hit the ground?
11
CHAPTER 1
1.2
Composite Functions
The picture on the right shows an oil leakage from a ship.
The oil leakage forms a circle. The circular area of the oil
leakage, A, is a function of its radius, r, in metres, and can be
modelled as A = f(r) = πr2.
The length of the radius, r increases with time, t,
in hours, measured from the moment the leakage starts.
This relationship can be modelled as r = g(t) = 100t.
By substituting r = 100t into the function A = f(r) = πr2,
we obtain:
A = f(100t)
= π(100t)2
= 10 000 πt2 m2
If the time t is given, then the area of the oil leakage can be determined. What can you say
about the combination of the two functions A = f(r) and r = g(t) which results in A = f [g(t)]?
Describing the outcome of composition of two functions
InQuirY 2
In groups
21st Century Learning
Aim: To explore the outcome of composition of two functions f and g
Instruction:
1. Scan the QR code or visit the link on the right.
2. Given the functions f(x) = x + 2 and g(x) = x2 together with the
bit.ly/2U5VrEq
respective graphs.
3. Examine the graphs formed on the plane.
4. Click on the f [g(x)] button and observe the graph displayed on the plane.
5. How to obtain the function f [g(x)]?
6. What is the shape of the graph resulted from the composition of the functions f and g?
7. Then, click again on the f [g(x)] button to delete the graph of f [g(x)].
8. Click on the g [f(x)] button and observe the graph displayed on the plane.
9. How to obtain the graph of g [f(x)]?
10. What is the shape of the graph resulted from the composition of the functions g and f ?
11. Then, change the functions f and g each with a different function to continue
exploring the results of the composition of two functions and their graphs.
12. Each group will move to the other groups to see the results.
13. Discuss with the members of your group regarding the results obtained by the others.
From the results of Inquiry 2, it was found that the function f[g(x)] is obtained by substituting
the function g into the function f whereas the function g[f(x)] is obtained by substituting the
function f into the function g.
12
1.2.1
The process of combination by replacing two functions f and g to generate f[g(x)] or g[f(x)] is
known as the composition of two functions and is written as fg(x) or gf(x).
fg(x) is read as “f composed with g of x” and is defined by fg(x) = f [g(x)].
Given two functions f(x) and g(x), the product of combination of two functions that
written as fg(x) or gf(x) are defined by fg(x) = f [g(x)] or gf(x) = g[f(x)].
Determining composite functions
Given functions f(x) = x + 2 and g(x) = x2. The diagram below shows part of the mapping of
function g followed by function f.
x
1
2
g
x2
y
1
4
f
y+2
z
3
1
2
x
g
f
➤
12 = 1
➤
1+2=3
➤
2 =4
➤
4+2=6
➤
x2 = y
➤
y + 2 = z = x2 + 2
g
g
2
f
f
6
fg
Based on the pattern in the diagram above, we can
simplify it into an arrow diagram as shown on the right.
From the arrow diagram, a direct mapping can be
seen with an element x  X mapped to an element z  Z
which can be defined by the function fg(x) = x2 + 2.
x
The new function of combining two functions f
and g with domain X and codomain Z is known as the
composite function of f and g which is represented by a
function fg.
Thus, from the process shown, we can conclude that:
X
fg(x) = f [g(x)]
x
Algebraically, the composite function fg(x) can be
determined as follows:
f(x) = x + 2
g(x) = x2
fg(x) = f [g(x)]
2
= f(x )
= x2 + 2 or fg : x → x2 + 2
g
f
y=x
2
z=y+2
= x2 + 2
Z
Y
fg
g
X
y = g(x)
f
z = f(y)
= f [g(x)]
Z
Y
x
1
2
fg
x2 + 2
3
6
1.2.1
1.2.2
13
CHAPTER 1
Functions
CHAPTER 1
Example 8
Two functions are defined by f : x → 2x and g : x → x2 – 5.
Determine the following composite functions.
(a) fg
(b) gf
(d) g2
(c) f 2
Solution
(a) fg(x) = f [g(x)]
(b) gf(x) = g[f(x)]
= g(2x)
= f(x2 – 5)
= (2x)2 – 5
= 2(x2 – 5)
= 4x2 – 5
= 2x2 – 10 2
Thus, fg : x → 2x – 10 Thus, gf : x → 4x2 – 5
Mind Challenge
Would the composite
functions, fg and gf always
be different?
MATHEMATICS
POCKET
f 2 is equal to ff.
Similarly, g 2 is equal to gg.
(d) g2 = g[g(x)]
(c) f 2(x) = f [ f(x)]
= f(2x)= g(x2 – 5)
= 2(2x) = (x2 – 5)2 – 5
= 4x
= x4 – 10x2 + 25 – 5
2
= x4 – 10x2 + 20
Thus, f : x → 4x
Thus, g2 : x → x4 – 10x2 + 20
Self Practice 1.4
1. In the arrow diagram on the right, function f maps set P to set Q
and function g maps set Q to set R.
Determine
(a) function f,
(b) function gf.
f
g
x
3x
2x – 7
P
Q
R
2. For each pair of the functions below, obtain an expression in the form of function notation for
fg, gf, f 2 and g2.
(a) f : x → 3x, g : x → 3 – x
(b) f : x → 4 + 2x, g : x → x2
(c) f : x → x + 4, g : x → 6 , x ≠ 0
x
(d) f : x → x – 5, g : x → 1 , x ≠ 1
x–1
3. Two functions, f and g are defined by f : x → 3x + 4 and g : x → x2 + 6. Find the expressions
for fg and gf, then find the values of x when
(a) f = g
(b) fg = gf
4. Given that f : x → ax + b and f 2 : x → 4x – 9, find the value of the constants a and b.
5. If f : x → 3x + k and g : x → 2h – 3x such that fg = gf, find the relation between h and k.
14
1.2.2
Determining the image or object of a composite function
By substituting the value of the object into a composite function, the image can be obtained.
Similarly, if the value of the image is given, then the value of the object can be determined by
solving the equation.
Example 9
If f : x → x – 1 and g : x → x2 – 3x + 4, find
(a) fg(2) and gf(1),
(b) the values of x when fg(x) = 7.
Solution
(a) fg(x) = f [g(x)]
(b)
fg(x) = 7
= f(x2 – 3x + 4) x2 – 3x + 3 = 7
= x2 – 3x + 4 – 1 x2 – 3x – 4 = 0
= x2 – 3x + 3 (x + 1)(x – 4) = 0
x = –1 or x = 4
Thus, fg(2) = (2)2 – 3(2) + 3
= 1 Thus, the values of x are –1 and 4.
gf(x) = g[ f(x)]
Alternative Method
= g(x – 1)
(a) g(2) = 22 – 3(2) + 4
= (x – 1)2 – 3(x – 1) + 4
2
=2
= x – 2x + 1 – 3x + 3 + 4
Thus, fg(2) = f(2)
2
= x – 5x + 8
=2–1
Thus, gf(1) = (1)2 – 5(1) + 8
= 1
=4
Self Practice 1.5
1. Given two functions f and g.
x , x ≠ 1, find fg(3).
x–1
(b) f : x → 5x + 6 and g : x → 2x – 1, find gf – 1 .
5
x
+
1
6
(c) f : x →
, x ≠ 3 and g : x →
, x ≠ 2, find f 2(4) and g2 1 .
x–3
x–2
2
2
(d) f : x → x2 – 4 and g : x →
, x ≠ 2, find f 2(–1) and g2(1).
x–2
(a) f : x → 2x + 1 and g : x →
( )
()
2. For each of the following function, find the value of the object x.
(a) f : x → 2x – 5, g : x → 10, x ≠ 0 and fg(x) = 5.
x
(b) f : x → x2 – 1, g : x → 2x + 1 and gf (x) = 7.
(c) f : x → 3x – 2 and f 2(x) = 10.
(d) g : x → 2 , x ≠ 2 and g2(x) = – 1 .
x–2
2
1.2.3
15
CHAPTER 1
Functions
CHAPTER 1
Determining a function when the composite function and one of the
functions are given
When the composite function and one of the functions are given, the other function can be
determined.
Example 10
The function f is defined by f : x → x – 2. Find the function g for each of the following.
(a) fg : x → 8x – 7
(b) gf : x → x2 + 3x – 5
Solution
(a)f [g(x)] = 8x – 7
(b) g[ f(x)] = x2 + 3x – 5
g(x) – 2 = 8x – 7
g(x – 2) = x2 + 3x – 5
g(x) = 8x – 7 + 2 Suppose y = x – 2
g(x) = 8x – 5
x=y+2
Thus, g : x → 8x – 5 Then, g(y) = (y + 2)2 + 3(y + 2) – 5
= y2 + 4y + 4 + 3y + 6 – 5
= y2 + 7y + 5
Replacing y with x, g(x) = x2 + 7x + 5
Thus, g : x → x2 + 7x + 5
Self Practice 1.6
1. Given the function f and the composite function fg, determine the function g for each of the
following.
(b) f : x → x2 + 1, fg : x → x2 + 4x + 5
(a) f : x → x – 3, fg : x → 2x2 – 4x + 7
2. Given the function f and the composite function gf, determine the function g for each of the
following.
(a) f : x → x + 1, gf : x → x2 – 2x – 3
(b) f : x → x2 + 3, gf : x → 2x2 + 3
3. Given the function h(x) = 8 , x ≠ 0 and hg(x) = 4x, find
x
(a) g(x),
(b) the value of x when gh(x) = 6.
4. Given the function g(x) = 3x and fg(x) = 9x – 7, find
(a) f(x),
(b) gf (2).
Solving problems involving composite functions
Example 11
Function f is defined by f : x → 12 , x ≠ 0.
x
(a) Express f 2(x), f 3(x) and f 4(x) in the simplest form.
(b) Hence, find f 22(x) and f 33(x).
16
1.2.4
1.2.5
Functions
(a) f(x) = 12 = x–2 x
f 2(x) = f [ f(x)]
f 3(x) = f [f 2(x)]
= f (x4)
= f 12
x
= 14 2
1
(x )
=
1 2
= 18
x2
x
3
= x4
= x –2
2
2
=x
CHAPTER 1
Solution
f 4(x) = f [f 3(x)]
= f 18
x
1
=
1 2
x8
= x16
4
= x2
( )
()
( )
( )
n
(b) From the pattern in (a), we can deduce that f n(x) = x–2 when n is odd and f n(x) = x2 when
22
33
n is even. Thus, f 22(x) = x2 and f 33(x) = x –2 .
Example 12
Mathematics
n
Application
Total production of q goods per day by a factory depends
on the number of workers, n, and the function is modelled
1
by q(n) = 10n – n2. Total revenue per day, r, in RM,
4
received from the sale of q goods is modelled by the
function r(q) = 40q. Determine the total revenue of the
factory in one day if the number of workers is 20.
Solution
1 . Understanding the problem
◆
1
Given two functions, q and r which are defined by q(n) = 10n – n2 and r(q) = 40q
4
respectively.
◆
Find the total revenue of the factory of 20 workers in one day.
2 . Planning the strategy
◆
◆
1.2.5
Find the composite function rq(n) so as to determine the total revenue of the factory, r
which represents the function of n workers, that is r(n).
Substitute n = 20 into the composite function r(n) which is obtained earlier so as to find
the total revenue of the factory per day, in RM.
17
CHAPTER 1
3 . Implementing the strategy
rq(n) = r[q(n)]
(
)
= r 10n – 1 n2
4
= 40 10n – 1 n2
4
= 400n – 10n2
Therefore, r(n) = 400n – 10n2
(
)
With 20 workers,
r(20) = 400(20) – 10(202)
= 8 000 – 4 000
= 4 000
Thus, the daily revenue of the factory
with 20 workers is RM4 000.
4 . Making a conclusion
When r(n) = 4 000,
4 000 = 400n – 10n2
2
10n – 400n + 4 000 = 0
n2 – 40n + 400 = 0
(n – 20)(n – 20) = 0
n = 20
Thus, the daily revenue of the factory is
RM4 000 when the number of workers
is 20.
Self Practice 1.7
x , x ≠ –1.
x+1
(a) Find the iterated functions f 2, f 3 and f 4. (b) Hence, write the functions f 20 and f 23.
1. The function f is defined by f : x →
2. If f : x → 1 , x ≠ 0, find
x
(a) the iterated functions f 2, f 3 and f 4,
(b) values of f 40(2) and f 43(2).
3. The surface area of a hot air balloon, A, in m2, filled with hot air is
given by the function A(r) = 4πr2 where r is the radius of the balloon,
in metres. The radius of the balloon is increasing as a function of
time, t, in seconds, according to the formula r(t) = 2 t³, t  0.
3
(a) State the surface area of the balloon, A, as a function of time, t.
(b) Find the surface area of the balloon after 2 seconds.
4. A cylindrical container of radius 20 cm contains 200 cm3 of water.
The container is filled with water at a constant rate of 100 cm3 per second.
(a) Write the formula of
(i) the amount of water in the container, v, after t seconds,
(ii) the height of water in the container, h, in terms of v,
(iii) the composite function hv(t).
(b) Find the height of water in the container after 20 seconds.
5. A small stone is thrown into a calm pond and produces a circular ripple. The radius of the
ripple, r, in cm, is increasing at a rate of 3 cm per second.
(a) Find an expression for the radius, r, in terms of time, t, after the stone is thrown.
(b) If A is the area of the ripple, explain the meaning of the composite function Ar(t).
(c) Find the area A, of the ripple after 30 seconds.
18
1.2.5
Intensive Practice 1.2
Scan the QR code or visit bit.ly/2AWCFIn for the quiz
1. Two functions are defined by f : x → 2x – 1 and g : x →
( )
(b) fg(2) and gf – 1 ,
2
(a) fg and gf,
CHAPTER 1
Functions
2. The functions f and g are defined by f : x →
x , x ≠ –1. Find
x+1
(c) the value of x when fg = gf.
x , x ≠ 1 and g : x → hx + k, where h and k are
x–1
constants. Given g(3) = 8 and gf(2) = 5, find
(b) the value of a if fg(a) = 3.
(a) the value of h and of k,
3. The functions f and g are defined by f : x → ax – b where a and b are constants and g : x → x + 4.
Given fg(2) = 9 and gf 1 = 2, find the value of a and of b.
2
4. The functions f and g are defined by f : x → 2 , x ≠ 3 and g : x → hx2 + k, where h and k
x–3
are constants.
(a) Given g(2) = 5 and gf(1) = –1, calculate the value of h and of k.
(b) Find the expression of gf.
()
5. Given that f : x → ax + b and f 3 : x → 27x + 13, find
(a) the value of a and of b,
(b) the expression of f 4.
6. The diagram on the right shows a tissue box with a square base of
side x cm and the height of 10 cm.
(a) Write the base area of the box, A as a function of x and its
volume, V as a function of A.
(b) Show that the volume, V is the result of the composition of
these two functions.
10 cm
x cm
x cm
7. The function f is defined by f : x → x + 6. Find the function g in each of the following.
(b) gf : x → x2 + 4
(c) gf : x → 8 – x
(a) fg : x → 2x2 – 3x – 7
➤
Q
➤
8. The diagram on the right shows the relation between set P, set Q
P
and set R. Given that set P maps to set Q by the function x – 1
3
and maps to set R by fg : x → x2 – 3x + 6.
fg(x) = x2 – 3x + 6
(a) Write the function that maps set P to set Q by using the
function notation.
R
(b) Find the function that maps set Q to set R.
➤
9. Given f : x → px + q and f 3 : x → 8x – 7,
(a) find the value of p and of q,
(b) determine the function f 4,
(c) by studying the pattern of f, f 2, f 3 and f 4, determine the general rule f n where n is the
number of times.
10. A car factory manufactures N cars daily after t hours of operation is given by N(t) = 100t – 5t 2,
0 < t < 10. If the cost, in RM, for manufacturing x cars is C(N) = 15 000 + 8 000x, find the cost
C as a function of time t, for the operation of the factory.
19
CHAPTER 1
1.3
Inverse Functions
You read news online that the temperature in New York is 39°F.
Calculate the temperature in degree Celsius.
The relationship between the number on a Fahrenheit, F
thermometer and that of degree Celsius, C is a function
9
F(C) = C + 32. By changing C as the subject of the formula,
5
5
that is C(F) = (F – 32) and substituting the value F = 39 into the
9
function C, the temperature in degree Celsius of New York
can be known.
F : C → 9 C + 32
5
C : F → 5 (F – 32)
9
(Source: https://www.necn.com/
weather/maps/ NECN-WeatherNow-250228521.html)
Will carrying out the inverse operation as shown above generate an inverse function of F?
Describing the inverse of a function
Inverse function of a function f can be written as f –1. For example:
f: x → x + 2
–1
f : x → x – 2
What is an inverse of a function? To understand this further, let’s follow the next exploration.
InQuirY 3
In groups
21st Century Learning
Aim: To explore the relation between the graph of a function and its
respective inverse.
Instruction:
1. Scan the QR code or visit the link on the right.
2. Click on the buttons of all the functions and study the graphs obtained. bit.ly/2LQjG8M
3. Are the graphs of each function and its inverse symmetrical about the line h(x) = x?
4. Carry out the discussion in the respective groups.
From the result of Inquiry 3, it was found that every graph of the function
and its graph of inverse function is symmetrical about the line h(x) = x, that
is y = x. The graph of f –1 is the reflection of the graph of f in the line y = x.
f : x → y ⇔ f –1 : y → x or y = f (x) ⇔ x = f –1(y)
20
f
x
y
f
–1
1.3.1
Example 13
f
x
In the arrow diagram on the right, the
function f maps x to y. Determine
(a) f –1 1
(b) f –1(5)
4
()
2
3
y
1
4
5
Solution
(a) From the given arrow diagram, we obtain
f(2) = 1 , thus f –1 1 = 2.
4
4
(b) By inverse mapping, f –1 : 5 → 3.
Then, f –1(5) = 3.
f : x → y ⇔ f –1: y → x
()
Example 14
A function is defined as f(x) =
(a) the image of 2 under f,
MATHEMATICS
POCKET
The sign –1 used in f –1 does
not mean the reciprocal of
f, f –1(x) ≠
1
but f –1 is
f(x)
the inverse of f.
x , x ≠ 4. Determine
x–4
(b) f –1(3).
Solution
(a) The image of 2, f(2) =
(b) Let a = f –1(3),
f(a) = 3
a = 3
a–4
a = 3(a – 4)
a = 3a – 12
2a = 12
a = 6
Thus, f – 1(3) = a = 6
2 = –1
2–4
f
3
a
f
–1
Self Practice 1.8
1. In the arrow diagram on the right, the function f maps x to y. Find
(a) f(4)
(b) f –1(–1)
(c) f –1(2)
(d) f –1(–5)
5
2. The functions g and h are defined by g(x) =
, x ≠ 2 and
2–x
h(x) = 3x + 6 respectively, find
(a) g(12)
1.3.1
(b) g–1(4)
(c) h(–1)
6
x
4
–2
f
y
2
–1
–5
(d) h–1(9)
21
CHAPTER 1
Functions
CHAPTER 1
Making and verifying conjectures related to the properties of inverse
functions
Do the following Inquiry 4, 5, 6 and 7 to make and verify the conjectures regarding the
properties of inverse functions.
InQuirY 4
In groups
21st Century Learning
Aim: To make and verify conjectures that a one-to-one function has an inverse function
Instruction:
1. Copy and complete the mapping of the following discrete functions.
(a) Discrete function g maps set A to set B and g –1 maps set B to set A.
x
g
–1
–2
l
2
A
g–1
x2
1
l
4
4
B
–1
–2
l
2
B
A
(b) The discrete function h maps set C to set D and h–1 maps set D to set C.
x
h
h–1
x+2
–2
0
2
4
0
2
4
6
0
2
4
6
–2
0
2
4
C
D
D
C
2. Are g –1 and h–1 functions?
3. What type of function will yield an inverse function? State your conjecture.
4. Every group must send a representative to present the outcome of the group’s
findings in front of the class. Members of the other groups can ask the representative
questions.
5. Repeat step 4 until all the groups have completed the presentation.
From the results of Inquiry 4, it was found that the inverse function is the reversal of a function
such that every element in the codomain is mapped to only one element in the domain. Thus,
we can conclude that:
A function f that maps set X to set Y has an inverse function, f –1
if f is a one-to-one function.
22
1.3.2
InQuirY 5
In groups
21st Century Learning
Aim: To make and verify conjectures that composite function
fg(x) = gf(x) = x such that function f and function g are
inverses of each other
Instruction:
1. The arrow diagram on the right shows the discrete function
f that maps set X to set Y and the discrete function g that
maps set Y to set X.
2. Complete the empty boxes below based on the arrow
diagram on the right.
f(1) = 1
f(2) = 3
f(3) = 0
f(0) = 0
g(1) = 1
g(3) = 2
g(5) = 0
g(0) = 0
f
1
2
3
4
1
3
5
7
X
Y
g
gf (1) = g(1) = 1
fg(1) = f(1) = 1
gf (3) = g(0) = 0
fg(5) = f(0) = 0
gf (2) = g(3) = 2
fg(3) = f(2) = 3
gf (4) = g(0) = 0
fg(7) = f(0) = 0
3. From the result obtained, what is the conjecture that can be made regarding the values
of fg(x) and gf(x)?
4. Each group presents their findings in front of the class and a question and answer
session is carried out.
From the results of Inquiry 5, the functions f and g are the inverse functions of each other if
and only if:
fg(x) = x, x in the domain of g and gf(x) = x, x in the domain of f.
InQuirY 6
In groups
21st Century Learning
Aim: To make and verify conjectures that if two functions f and g are
inverse functions of each other, then
(a) domain of f = range of g and domain of g = range of f
(b) graph of g is the reflection of graph of f at the line y = x
Instruction:
bit.ly/2ob10HU
1. Scan the QR code or visit the link on the right.
2. Click on the box f(x) = 1 x for the domain 0 < x < 8 and take note of the graph obtained.
2
3. Then, click on the box g(x) = 2x, that is the inverse of f and take note of the graph
displayed.
4. Complete the domain and range of the
Graph
Domain
Range
graphs of f and g in the table shown on
Graph of function f
the right.
5. What is your conjecture regarding the
Graph of function g
results of the findings?
6. How are the graphs of f and g located with respect to the line y = x? What is your
conjecture?
7. Each representative of the groups moves to the other groups and presents the results
of their group.
1.3.2
23
CHAPTER 1
Functions
CHAPTER 1
From the results of Inquiry 6, it was found that:
If two functions f and g are inverse functions of each other, then
(a) the domain of f = range of g and domain of g = range of f
(b) the graph of g is the reflection of the graph of f at the line y = x
InQuirY 7
In groups
21st Century Learning
Aim: To make and verify conjecture that if the point (a, b) is on the
graph of f, then the point (b, a) is on the graph of g
Instruction:
bit.ly/30PXAr9
1. Scan the QR code or visit the link on the right.
2
2. Click on the box f(x) = x + 1 in the range 0 < x < 3 and its inverse
function, g(x) = √x – 1 in the range 1 < x < 10.
3. Then, click on the box “Point and reflection”. Drag point A along the graph f. Note the
points on the graph f and the graph g.
4. What is the conjecture that can be made regarding the points that you observe on
both graphs?
5. Carry out a discussion within the group regarding the results of the finding.
6. Each group elects a representative and presents in front of the class.
From the results of Inquiry 7, it was found that:
For any real number, a and b, if the point (a, b) lies on the graph f, then the point
(b, a) lies on the graph g, that is graph f –1. The point (b, a) lies on the graph g is
the point of reflection of (a, b) which lies on the graph f in the line y = x.
To determine whether the graph of a function has an inverse function, carry out the
horizontal line test. If the horizontal line cuts the graph of the function at only one point, then
this type of function is a one-to-one function and it has an inverse function. Conversely, if the
horizontal line cuts the graph at two or more points, then this type of function is not a one-toone and the function has no inverse function.
y
f(x)
0
y
y = f(x)
x
f has an inverse function
24
f(x)
Horizontal
line test
x
y = f(x)
0
Horizontal
line test
x1
x2
x
f does not have an inverse function
1.3.2
Example 15
Determine whether each of the following functions f has an inverse or not. Give a reason for
your answer.
y
y
(a)(b)(c)
f
x
y
a
–2
b
4
c
9
(3, 7)
f
f
0
(–3, –3)
Domain: –3 x 3
x
0
–4
x
4
Domain: –4 x 4
Solution
(a) f is a function because this function is one-to-one with every
element in the domain mapped to only one element in the
codomain. The inverse of this function also maps every
element in the codomain to only one element in the domain.
Thus, function f has an inverse function.
y
(b) When the horizontal line test is carried out, the horizontal
line cuts the graph of function f at only one point. This
(–3, –3)
means that function f is a one-to-one function. Thus,
function f has an inverse function.
(c) When the horizontal line test is carried out, the horizontal
line cuts the graph of function f at two points. This means
that function f is not a one-to-one function. Thus, function
f has no inverse function.
f
(3, 7)
Horizontal
line test
x
0
y
f
–4
Horizontal
line test
0
4
x
Example 16
Verify that the function f(x) = 3 – 2x has an inverse function, g(x) = 3 – x.
2
Solution
First, determine fg(x).
fg(x) = f [g(x)]
=f 3–x
2
=3–23–x
2
= 3 – (3 – x)
=x
( )
( )
Since fg(x) = gf(x) = x, thus g(x) =
1.3.2
Then, determine gf(x).
gf(x) = g[ f(x)]
= g(3 – 2x)
3 – (3 – 2x)
=
2
= 2x
2
=x
3–x
is the inverse function of f(x) = 3 – 2x.
2
25
CHAPTER 1
Functions
The function f is defined by f : x → x2 – 6 for the domain 0 < x < 4. On the same plane,
sketch the graphs of f and f –1. Hence, state the domain of f –1.
y
Solution
The graph of f is a part of the quadratic curve y = x2 – 6. By plotting the
points in the table of values below, the graph of f is sketched as shown
in the diagram on the right.
x
y
0
1
–6
–5
2
3
–2
3
(4, 10) y = x
10
(10, 4)
f –1
4
–6
0
x
f
–6
The graph of f –1 is the reflection of the graph of f in the line y = x.
The domain of f –1 is the range of f. Hence, the domain of f –1 is –6  x  10.
y
Example 18
Q(3, 6)
The diagram on the right shows the graph of y = f(x)
passing through the points P(1, 0) and Q(3, 6). On the same
diagram, sketch the graph of y = f –1(x) by showing the points
corresponding to point P and point Q.
The graph of y = f –1(x) is the reflection of the graph of y = f(x)
in the line y = x. The points P9 and Q9 on the graph of y = f –1(x)
correspond to the points P and Q as shown in the diagram on the
right.
y
x
0 P(1, 0)
Q(3, 6)
y=x
(x)
Solution
y = f(x)
y=f
CHAPTER 1
Example 17
P'(0, 1)
Q'(6, 3)
y = f (x)
–1
x
0 P(1, 0)
Self Practice 1.9
1. Determine whether each of the following function given has an inverse function.
g
y
(a)(b)(c)
f
x2 – 3
x
x
2+ 6
x
f(x) = x – 2
1
3
4
–4
2
3
6
4
5
8
(d) {(1, 2), (4, 5), (5, 8), (9, 9)}
(f) f : x → 4 – x2
26
3
4
6
13
–2 0
2
6
x
(e) {(–3, 2), (–1, 1), (2, 4), (5, 4), (9, 5)}
(g) f : x → 1 2, x > 2
(x – 2)
1.3.2
2. Are the following functions f and g the inverse functions of each other? Verify the truth by
applying the relation fg(x) = gf(x) = x.
(a) f(x) = 3x – 2 and g(x) = x + 2
3
2x
(b) f(x) =
, x ≠ 3 and g(x) = 3x , x ≠ 2
x–3
x–2
2
3x
, x ≠ 3 and g(x) = – 2, x ≠ 0
(c) f(x) =
x–3
x
(d) f(x) = 2 + 5x and g(x) = x – 5
2
3. The function f is defined as f : x → x3 for the domain –1 < x < 2. On the same plane, sketch
the graphs of f and f –1. Hence, state the domain and range of f –1.
4. The function h is defined as h(x) = x2 – 2 for the domain 0  x  3.
(a) On the same diagram, sketch the graphs of h and h–1.
(b) State the domain of h–1.
(c) Find the value of x such that h(x) = h –1(x).
5. The coordinates of the following points lie on the graph of one-to-one function, f. Determine
the corresponding coordinates lying on the graph of f –1.
(a) P –2, 1 (b) Q(1, –3) (c) R(4, 5) (d) S(–6, –8)
2
(
)
y
6. The diagram on the right shows the line y = x and the graph of
B(3, 10) y = x
y = f(x) for the domain –1 < x < 3. The points A(–1, 2), B(3, 10)
y = f(x)
and P(a, b) lie on the graph.
(a) Sketch the graph of y = f –1(x) that shows the points on the
P(a, b)
graph of y = f –1(x) corresponding to the points A and B.
(b) Find the values of a and b, if the corresponding coordinates A(–1, 2)
x
on the graph of y = f –1(x) are (4, 1).
0
Determining the inverse function
We have learnt that when given y = f(x), then x = f –1(y).
Algebraically, the formula of the inverse function, f –1­(x) with
the original function y = f (x) can be determined by following the
steps below.
f
x
y
f
Change function
y = f(x) to the form
x = f(y).
1.3.2
1.3.3
Write x as f –1­(y).
–1
Replace variable y
with variable x.
27
CHAPTER 1
Functions
CHAPTER 1
Example 19
If f : x → 5x + 2, find
(a) f –1(x)
MATHEMATICS
POCKET
(b) f –1(7)
Check the validity of the
inverse function
x–2
f –1(x) =
obtained in
5
Example 19(a) by applying
the relationship
ff –1(x) = f –1f(x) = x.
Solution
(a) f(x) = 5x + 2
Let
y = 5x + 2
5x = y – 2
y–2
The form of x = f(y)
x=
5
Since x = f –1(y), Write x as f –1(y)
f –1(y) = x
y–2
=
5
Substitute the variable y with x,
x–2
f –1(x) =
5
x–2
f
Thus, f –1 : x →
.
5
x–2
(b) f –1(x) =
1
5
7–2
Thus, f –1(7) =
5
–1
f
=1
ff –1(x) = f [f –1(x)]
x–2
+2
= 5
5
= x
f –1f(x) = f –1[f(x)]
= f –1(5x + 2)
5x + 2 – 2
=
5
= x
–1
Since ff (x) = f –1f(x) = x,
x–2
is the
thus f –1 : x →
5
inverse function of
f : x → 5x + 2.
( )
7
Self Practice 1.10
1. Find f –1 for each of the following one-to-one functions.
(a) f : x → 2x – 5
(b) f : x → 3 , x ≠ 0
x
5x
x
,x≠6
(e) f : x → + 9, x ≠ 8
(d) f : x →
x–6
x–8
(c) f : x → 4 , x ≠ 1
x–1
(f) f : x → 2x – 3, x ≠ 1
2x – 1
2
2. The function f is defined by f : x → 3 – x, x ≠ 0, find
2x
(a) f –1(4),
(b) the values of x such that f(x) = f –1(x).
5
, find the value of the constants a and b.
8
4. Find the function f in similar form for each of the following f –1.
(a) f –1 : x → 6x + 7
(b) f –1 : x → 2 – x
(c) f –1 : x → 3x , x ≠ 3
5
x–3
4
–1
–1
5. The inverse function g is defined by g : x →
, x ≠ k.
2–x
(a) State the value of k.
(b) Find g 1 .
2
1.3.3
28
3. Given the functions h : x → 4x + a and h–1 : x → 2bx +
()
Intensive Practice 1.3
CHAPTER 1
Functions
Scan the QR code or visit bit.ly/2p5bB7a for the quiz
1. In the arrow diagram on the right, the function f maps x to y and
the function g maps y to z. Determine
(b) g(5)
(a) f (2)
(d) f –1(5)
(c) gf (2)
(f) f –1g–1(8)
(e) g–1(8)
2
f
x
g
y
5
z
8
2. By applying the horizontal line test, determine whether each of the following function has an
inverse function.
(a)(b)(c)
y
y
y
1
–
x
0
1 3 – 5x
f(x) = –x
2
1
f(x) = –
x
f(x) = 2x 3
x
0
x
0
3. The diagrams below show the graphs of one-to-one functions, f. In each case, sketch the
graph of f –1 and hence state the domain of f –1.
y
y
y
(a)(b)(c)
(4, 8)
4
f
0
(4, 2)
f
f
x
0
2
x
0
x
4. Given f : x → 2x + h, x ≠ 3 and f(4) = 13, find
x–3
(a) the value of h,
(b) f –1(3),
–1
(c) the value of m when f (m) = 2.
2
5. The inverse function h–1 is defined by h–1 : x →
, x ≠ 3, find
3–x
(a) h(x),
(b) the values of x such that h(x) = 2.
5
1
6. Two functions f and g are defined by f : x → 4x – 17 and g : x →
, x ≠ 3 . Solve
2x – 7
2
the equation f –1(x) = g–1(x).
7. Faridah carried out a physical activity during her leisure time. Then, Faridah calculated her
estimated heartbeat rate by using the function f(x) = 17(220 – x), where x is her age.
20
(a) Determine the inverse function of this function.
(b) If Faridah’s age is 16, determine her estimated heartbeat rate.
29
Summary OF Chapter 1
Arrow diagram
o�
f : x → 2x or
f(x) = 2x,
x is the object
2x is the image
��
a
��
Graph
r��
r�
�e
��n
��i�
� ��
t� � i � � FUNCTIONS
Discrete functions
v�
�� ��v��
Continuous functions
One-to-one function, f : x → y has
an inverse function, f –1 : y → x
Composite functions
Inverse functions
n
��n � � t � �
�
CHAPTER 1
8. Zaki intends to make spherical water balls that can hold
1
cm3 of water. The volume of sphere, V is given by
2
4
V = πr3, where r is the radius of the sphere. Zaki
3
wishes to know how to determine r if V is given.
(a) Draw an arrow diagram of the function f that maps r
to V and its inverse function f –1 which maps V to r.
(b) Hence, determine the radius of the ball that can hold
the volume of water according to the specification.
gf(x) = g[f(x)]
x
gf
f
x
g
f(x)
g[f(x)]
fg(x) = f[g(x)]
fg
g
x
30
f
g(x)
f
f [g(x)]
y
f –1
Properties of inverse functions:
• Only restricted to one-to-one function
which has inverse function.
• fg(x) = gf(x) = x, where g(x) is the
inverse function of f(x).
• Graph of f –1 is the reflection of graph of f in the line y = x and vice versa.
• If the point (a, b) lies on the graph
of f, then the point (b, a) lies on the
graph of f –1.
CHAPTER 1
Functions
WRITE YOUR JOURNAL
Search on the internet or books regarding the history of the usage of the function notation
y = f(x). Create a digital folio by using presentation software such as PowerPoint, Prezi or
Powtoon.
MASTERY PRACTICE
1. The arrow diagram on the right shows the relationship between
set M and set N. PL1
(a) State
(i) the image of 2,
(ii) the object of 4.
(b) Does this relation represent a function? Give your reason.
(c) State the domain, codomain and the range of the relation.
2. The arrow diagram on the right shows a function h. PL2
(a) State the value of m.
(b) Using the function notation, express h in terms of x.
2
6
7
8
9
Set M
1
4
5
Set N
h
x
h (x)
–2
2
4
6
3. Using the vertical line test, determine whether the graph on
the right represents a function or not. If yes, is this function a
one-to-one function? Test by drawing a horizontal line on the
graph. PL2
3
15
m
y
x
0
4. Function f is defined by f : x → |x – 3| for the domain –1  x  7. PL3
(a) Sketch the graph of f and state the range of f.
(b) Find the range of values of x such that f(x) < 2.
(c) On the same graph in part (a), sketch the graph of y = 2x – 3 and hence obtain the value
of x such that |x – 3| = 2x – 3.
5. The arrow diagram on the right represents part of the mapping of
f : x → hx + k , x ≠ 0, find PL3
x
(a) the value of h and of k,
(b) the image of 6 under this mapping.
x
f
hx + k
x
2
3
17
23
6. Two functions f and g are defined by f : x → x + 2, x ≠ 2 and g : x → mx + c. Given that
x–2
g –1(2) = f(3) and gf –1(2) = 5, find the value of m and of c. PL3
31
CHAPTER 1
7. In the diagram on the right, the function f maps set A to set B and
the function g maps set B to set C. Find PL4
(a) in terms of x, the function
(i) that maps set B to set A,
(ii) g(x).
(b) the value of x such that fg(x) = 4x – 3.
x
f
A
8. Function f is defined by f : x → m + n, x ≠ k. Given f(2) = 3 and f(3) = 2, find
x–1
(a) the value of k,
(b) the value of m and of n,
(d) f –1(2).
(c) f 2(x),
3x – 2
g
6x + 1
B
C
PL3
y
9. The diagram on the right shows the function f(x) = –x3 – 3x2 – x + 1
for the domain –3  x  1. PL3
4
(a) State
(i) whether the function f is discrete or continuous,
1
(ii) the range of values of f corresponding to the domain
given.
0
–3
(b) By applying the horizontal line test, determine whether f
has an inverse function or not.
f(x) = –x3 – 3x2 – x + 1
1
x
–4
10. Given that the functions f(x) = |x| and f(x) = x4 are not one-to-one functions. PL5
(a) Determine the suitable conditions in the domain of f so that the new functions become
one-to-one functions.
(b) From (a), find the inverse function for each of the functions f.
11. If the graphs of a function and its inverse function intersect, would the two graphs intersect
on the line y = x? What is the possibility for the two graphs to intersect on other lines? PL5
12. Given f(x) = ax + b, find f –1(x). PL5
cx + d
(a) Using the formula f –1 obtained, determine f –1 for each of the following functions.
(i) f(x) = x + 8, x ≠ 5
x–5
2x
(ii) f(x) = – 3 , x ≠ – 4
x+4
(b) If c ≠ 0, what are the conditions on a, b, c and d so that f = f –1?
13. A one-to-one function f is defined by f : x → x2 – 2x for 1  x  3. PL6
(a) Using the GeoGebra software,
(i) draw the graph of f and from the graph, state the range of f,
(ii) draw the graph of f –1 on the same plane and state the domain of f –1.
(b) What can you say about the range of f and the domain of f –1 and also the domain of f and
the range of f –1? Hence, on the same plane, draw the line y = x.
(i) Is the graph of f –1 the reflection of the graph f in that line?
(ii) Is the point (0, 2) on the graph of f –1 the reflection of the point (2, 0) on the graph of
f in the line y = x? What conclusion can you make?
32
Functions
CHAPTER 1
14. The price p, in RM, of an item and the quantity x sold follow the demand equation
1
p = 100 – x for 0 < x < 400. Whereas the cost C, in RM, to produce x units is
4
x
+ 600. Assuming all the items produced are sold, calculate PL4
C=
25
(a) the cost C as a function of price p,
(b) the cost for producing that item if the price for one unit of the item is sold at RM36.
15. Period T, in seconds, of a simple pendulum is a function of length l, in metres, defined by
l , such that g = 10 m s–2 is the gravitational acceleration. Using the GeoGebra
T(l
g
software, draw the graph of this function and on the same plane, draw the graphs of the
following functions.
l+4
(a) T(l) = 2π w
g
4l
(b) T(l) = 2π w
g
!
!
How does the change in the length, affect the period, T of the pendulum?
Exploring
PL5
MATHEMATICS
The table below shows the amount of petrol used by a car on a highway as compared to the
distance travelled. Supposed l is the volume of petrol used, in litres, and d is the distance
travelled, in km, by the car.
Petrol used (l)
4
8
12
16
20
Distance travelled, in km (d)
48
96
144
192
240
1. Based on the table above,
(a) how far can the car travel with l litre of petrol?
(b) determine the distance travelled, d, as a function of amount of petrol used, l.
d(l) =
2. Using the GeoGebra software, draw the function d as obtained in question 1(b) and from
the graph, determine the following:
(a) What is the amount of petrol used to travel 300 km?
(b) What is the distance that can be travelled for 100 l of petrol?
33
Quadratic
Functions
CHAPTER 2
CHAPTER
2
What will be learnt?
Quadratic Equations and Inequalities
Types of Roots of Quadratic Equations
Quadratic Functions
List of
Learning
Standards
bit.ly/2AYDj8t
wORDS
KEY W
Completing the
square
Root
General form
Quadratic inequality
Number line
Discriminant
Real root
Imaginary root
Vertex form
Axis of symmetry
Maximum value
Minimum value
34
Penyempurnaan
kuasa dua
Punca
Bentuk am
Ketaksamaan
kuadratik
Garis nombor
Pembezalayan
Punca nyata
Punca khayalan
Bentuk verteks
Paksi simetri
Nilai maksimum
Nilai minimum
2.1.1
Quadratic Functions
Know?
A satellite dish has the ability to converge the
energy on its focal point. Satellite, television,
radar and telecommunication tower are
examples of objects which focus on the
properties of reflection of parabola.
Based on the history of the ancient
times, Archimedes helped the Greek army by
using parabolic mirrors to torch the military
ships of Rome who were trying to conquer the
Greek town, Syracuse in 213 B.C.
For further information:
bit.ly/35rNxMi
SIGNIFICANCE
OF THIS CH
APTER
Astronomers use the concept of quadratic
function in inventing telescopes. Parabolic
mirrors are able to converge and reflect the
light onto a certain point.
In the engineering field, engineers apply the
concept of quadratic function to determine the
types of loads which can be accommodated
by a bridge.
The cross section of a skateboard ramp in the
shape of a parabola can be modelled by using
the quadratic function, that is f(x) = ax2 + bx + c.
For your knowledge, the shape and width of a
skateboard ramp can be modified through the
knowledge of quadratic functions. What is the
best shape of skateboard ramp from the safety
aspects?
2.1.1
Scan this QR code
to watch a video on
skateboard games in
Malaysia.
bit.ly/2V2I1ys
35
CHAPTER 2
Did you
2.1
Quadratic Equations and Inequalities
CHAPTER 2
Solving quadratic equations by using the method of completing the
square and quadratic formula
Most of the situations that take place in our daily lives are
associated with equations. One of the equations is the quadratic
equation. Consider this situation:
The area of a rectangular picture frame is 100 cm­­2. If the length
is 3 cm longer than its width, write an equation which satisfies
this situation.
Supposed that the width of the frame is x cm and its length is 3 cm longer than its width, that is
(x + 3) cm. Then:
x(x + 3) = 100
x2 + 3x = 100
x2 + 3x – 100 = 0
Note that this equation has a variable x and the highest power of the variable is 2. Hence,
this equation is known as a quadratic equation in general form. In general, a general form of
quadratic equation can be written as:
ax2 + bx + c = 0 where a, b and c are constants and a ≠ 0.
How do you solve a quadratic equation? What does it mean by solving the quadratic
equation?
InQuirY 1
In pairs
21st Century Learning
Aim: To explore the solving of quadratic equations by using dynamic
geometry software
Instructions:
bit.ly/30MWj48
1. Scan the QR code or visit the link on the right.
2. Click the point button and mark A and B on the intersection
points between the graph y = 3x2 + 11x – 4 and the x-axis.
3. Record the coordinates of point A and point B. Then, observe the x-coordinate of point
A and point B.
4. What is the conclusion that can be made on the x-coordinate of point A and point B?
5. Discuss with your partner and share the findings obtained with other classmates.
From the results of Inquiry 1, the values of x for both the intersection points, that is, x = –4 and
1
are the solutions or roots of the equation y = 3x2 + 11x – 4 when y = 0.
x=
3
36
2.1.1
Quadratic Functions
Then, it can be concluded that:
You have learned how to solve quadratic equations by using factorisation method. Other than
that, the solutions of a quadratic equation can be obtained by using the completing the square
and formula method.
A Completing the square method
FLASHBACK
Example 1
Factorisation method
x2 + 5x + 6 = 0
(x + 2)(x + 3) = 0
x = –2 or x = –3
Solve the following equations by using completing the square
method.
(a) x2 + 4x – 7 = 0
(b) –3x2 + 6x – 1 = 0
Move the constant term to the
right hand side of the equation
Solution
(a) x2 + 4x – 7 = 0
2
coefficient of x
x2 + 4x = 7
Add the term
2
2
2
x2 + 4x + 4 = 7 + 4
on the left and right hand side
2
2
of the equation
x2 + 4x + 22 = 7 + 22
(x + a)2 = x2 + 2ax + a2
(x + 2)2 = 11
11
x + 2 = ±!w
x = –5.317 or x = 1.317
Hence, the solutions of equation x2 + 4x – 7 = 0 are –5.317
and 1.317.
()
(
()
(b)–3x2 + 6x – 1 = 0
x2 – 2x + 1 = 0
3
Divide both sides of the equation by –3
so that the coefficient of x2 becomes 1
x2 – 2x = – 1
3
2
2
–2
1
= – + –2
x2 – 2x +
2
3
2
1
x2 – 2x + (–1)2 = – + (–1)2
3
2
(x – 1)2 =
3
2
x – 1= ± w
3
( )
( )
!
Add
Factorisation method by
using algebra tiles.
bit.ly/34f7x3G
Mathematics Museum
(–22) on both sides of the
2
equation
x = 0.1835 or x = 1.8165
Hence, the solutions of equation –3x2 + 6x – 1 = 0 are
0.1835 and 1.8165.
2.1.1
)
QR
A Persian mathematician,
Abu Ja’far Muhammad ibn
Musa al-Khawarizmi used the
same method as completing
the square to solve quadratic
equations.
37
CHAPTER 2
Solutions or roots of the quadratic equation ax2 + bx + c = 0 are the x-coordinates of the
intersection points between the graph y = ax2 + bx + c and the x-axis.
B Formula method
Mind Challenge
CHAPTER 2
The formula for solving a quadratic equation ax2 + bx + c = 0 is
given as:
x=
Derive the quadratic formula
by using completing the
square method.
–b ± !w
b2 – 4ac
2a
BRAINSTORMING
State another method to
solve a quadratic equation
other than the method of
completing the square and
quadratic formula. What is
your choice? Explain the
reason for your choice.
Example 2
Solve the equation 2x2 – 2x – 3 = 0 by using formula.
Solution
Compare the given equation with the equation of general form
ax2 + bx + c = 0. Hence, a = 2, b = –2 and c = –3.
–(–2) ± !w
(–2)2 – 4(2)(–3)
2(2)
2 ± !w
28
=
4
2 – !w
28
2 + !w
28
or x =
x=
4
4
= –0.823 or
= 1.823
x=
Mathematics Museum
Hence, the solutions of the equation 2x2 – 2x – 3 = 0 are –0.823
and 1.823.
An Indian mathematician
and astronomer,
Brahmagupta produced a
formula to solve a quadratic
equation ax2 + bx + c = 0
which is equivalent to
x=
4ac + b2 – b
!w
2a
.
Self Practice 2.1
1. Solve the following quadratic equations by using completing the square method. Give your
answers in three decimal places.
(a) x2 + 4x – 9 = 0
(b) x2 – 3x – 5 = 0
(c) –x2 – 6x + 9 = 0
2
2
(e) 4x – 8x + 1 = 0
(f) –2x2 + 7x + 6 = 0
(d) 2x – 6x + 3 = 0
2. Solve the following quadratic equations by using formula. Give your answers in three
decimal places.
(b) 2x2 + 2x – 1 = 0
(c) 3x2 – 8x + 1 = 0
(a) x2 – 4x – 7 = 0
2
(e) (x – 1)(x – 3) = 5
(f) (2x – 3)2 = 6
(d) 4x – 3x – 2 = 0
3. (a) The length of the diagonal of a rectangle is 10 cm. If the length is 2 cm longer than its width, find the length and the width of the rectangle.
(b) Find the measurements of a rectangle with a perimeter of 26 cm and an area of 40 cm2.
4. The diagram on the right shows a trapezium PQRS where
PQ = (x – 1) cm, PS = (x + 3) cm and QR = (3x + 2) cm.
Given the area of the trapezium is 17 cm2, find the value
of x.
38
P
(x + 3) cm
S
(x – 1) cm
Q
(3x + 2) cm
R
2.1.1
Quadratic Functions
Forming quadratic equations from given roots
Comparing
1
and
2
,
– (α + β) = b
a
and
α+β=–b
a
FLASHBACK
Factorisation identity
(a) (x + y)2 = (x + y)(x + y)
= x2 + 2xy + y2
(b) (x – y)2 = (x – y)(x – y)
= x2 – 2xy + y2
(c) x2 – y2 = (x + y)(x – y)
αβ = c
a
BRAINSTORMING
In general, this comparison can be formulated as follows:
Sum of roots = α + β = – b
a
c
Product of roots = αβ =
a
Therefore, the quadratic equation with roots a and b can be
written as:
x2 – (sum of roots)x + (product of roots) = 0
Given α =
–b – !w
b2 – 4ac
2a
–b + !w
b2 – 4ac
,
2a
(a) show that
α+β=– b,
a
(b) express the product of
αβ in terms of a and c.
and β =
Discuss with your
classmates.
Example 3
Form a quadratic equation with roots 3 and –5.
Solution
Given α = 3 dan β = –5.
Sum of roots, α + β = 3 + (–5)
= –2
Product of roots, αβ = 3 × (–5)
= –15
Thus, the quadratic equations with roots 3 and –5 is
x2 – (sum of roots)x + (product of roots)
x2 – (α + β)x + αβ
x2 – (–2)x + (–15)
x2 + 2x – 15
2.1.2
=0
=0
=0
=0
Alternative Method
(x – 3)(x + 5) = 0
x2 + 5x – 3x – 15 = 0
x2 + 2x – 15 = 0
39
CHAPTER 2
The quadratic equation ax2 + bx + c = 0 can be written as
x2 + b x + c = 0 … 1
a
a
If a and b are the roots of a quadratic equation, then
(x – α)(x – β) = 0
x2 – (α + β)x + αβ = 0 … 2
CHAPTER 2
Example 4
If a and b are the roots of the quadratic equation 2x2 + x = 4, form a quadratic equation with
the following roots.
(a) α + 3, β + 3
(b) 2α, 2β
(c) α2, β2
Solution
2x2 + x – 4 = 0 where a = 2, b = 1 and c = – 4
α + β = – b = – 1 and αβ = c = – 4 = –2
a
2
a
2
Product of roots:
(a) Sum of roots:
(α + 3)(β + 3) = αβ + 3(α + β) + 9
(α + 3) + (β + 3) = (α + β) + 6
= –2 + 3 – 1 + 9
=– 1 +6
2
2
11
=
= 11
2
2
Thus, the quadratic equation with roots a + 3 and b + 3 is
Multiply both sides of the
x2 – 11x + 11 = 0
2
2
equation by 2
2x2 – 11x + 11 = 0
( )
Product of roots:
(b) Sum of roots:
(2α)(2β) = 4αβ
2α + 2β = 2(α + β)
= 4(–2)
=2– 1
= –8
2
= –1
Thus, the quadratic equation with roots 2a and 2b is
x2 – (–1)x – 8 = 0
x2 + x – 8 = 0
( )
Product of roots:
(c) Sum of roots:
2
2
2
α2β2 = (αβ)2
α + β = (α + β) – 2αβ
2
= (–2)2
= – 1 – 2(–2)
=4
2
1
= + 4
4
= 17
4
Thus, the quadratic equation with roots a2 and b2 is
Multiply both sides of the
x2 – 17 x + 4 = 0
4
equation by 4
4x2 – 17x + 16 = 0
( )
40
2.1.2
Quadratic Functions
Self Practice 2.2
(a) 2 and 6
(c) −4 and −7
(b) −1 and 4
(d) 1 and −5
5
2. The quadratic equation x2 + (p – 5)x + 2q = 0 has roots of –3 and 6. Find the value of p and of q.
3. If a and b are the roots of the quadratic equation 5x2 – 10x – 9 = 0, form quadratic equations
with the following roots.
β
(a) α + 2 and β + 2 (b) 5α and 5β
(c) α – 1 and β – 1
(d) α and
3
3
4. If a and b are the roots of the quadratic equation 2x2 + 5x = 1, find the quadratic equations
with the following roots.
β
1 1
1
1
(a) ,
(c) α2, β 2
(d) α ,
(b) α + , β +
α
α β
α
β
β
(
)(
)
5. A quadratic equation 2x2 = 6x + 3 has two roots p and q. Find the quadratic equations with
roots p2q and pq2.
Solving quadratic inequalities
An inequality with a quadratic expression on one side and zero on the other side, is called a
quadratic inequality in one variable. For example, 2x2 + 7x – 4 < 0 and (x + 1)((x – 3) . 0 are
quadratic inequalities in one variable, x. To solve a quadratic inequality as (x + 1)(x – 3) . 0,
we have to find the range of values of x so that the expression on the left is greater than zero.
The three methods which can be used to solve a quadratic inequality are graph sketching,
number line and table methods.
InQuirY 2
In groups
21st Century Learning
Aim: To solve quadratic inequalities by graph sketching, number line and table methods
Instructions:
1. Consider quadratic inqualities (x + 1)(x − 3) . 0 and (x + 1)(x − 3) , 0.
2. Form three groups and each group has to choose one of the three following method.
Graph sketching method
✑ Solve the quadratic equations (x + 1)(x − 3) = 0.
✑ Draw the graph of y = (x + 1)(x − 3).
✑ Mark and determine the range of values of x on the graph when (x + 1)(x − 3) . 0
(y . 0) and (x + 1)(x − 3) , 0 (y , 0).
2.1.2
2.1.3
41
CHAPTER 2
1. Form quadratic equations which have the following roots.
Number line method
CHAPTER 2
✑ Solve the quadratic equation (x + 1)(x − 3) = 0.
✑ Draw a number line on a piece of paper.
✑ By choosing the values of x that satisfy x , −1, x . 3 and −1 , x , 3 on the
number line and substituting them into (x + 1)(x − 3), determine and verify the
range of values of x when (x + 1)(x − 3) . 0 and (x + 1)(x − 3) < 0.
Table method
✑ Copy and complete the table with positive value (+) or negative value (–) for
every factor of quadratic equation (x + 1)(x – 3) = 0.
x < –1
x = –1
–1 < x < 3
x=3
x>3
(x + 1)
(x – 3)
(x + 1)(x – 3)
✑ From the results obtained in the table, what is the range of values of x when
(x + 1)(x − 3) . 0 and (x + 1)(x − 3) , 0?
3. Compare the findings of your group with other groups.
4. Do a thorough discussion about the three methods that can be used to solve quadratic
equations.
From the results of Inquiry 2, the solution for the quadratic inequalities (x + 1)(x – 3) . 0 and
(x + 1)(x – 3) , 0 by using the methods of graph sketching, number lines and table are shown
as follow:
Graph sketching
Number line
Test point –2:
Test point 0:
(–2 + 1)(–2 – 3) > 0 (0 + 1)(0 – 3) < 0
y>0
y
x>3
x < –1
–1 0
y<0
–1 < x < 3
+
y>0
3
x
–
x < –1 –1 –1 < x < 3 3
Test point 4:
(4 + 1)(4 – 3) > 0
+
x>3
x
Table
(x + 1)
(x – 3)
(x + 1)(x – 3)
Range of values of x
x < –1
–1 < x < 3
x>3
–
+
+
–
–
+
+
–
+
From the three findings above, it can be concluded that:
For a quadratic equation in the form of (x − a)(x − b) = 0, where a < b,
(a) if (x − a)(x − b) > 0, then x < a or x > b,
(b) if (x − a)(x − b) < 0, then a < x < b.
42
2.1.3
Quadratic Functions
Example 5
CHAPTER 2
Find the range of values of x for the quadratic inequality (2x – 1)(x + 4)  x + 4 by using
(a) graph sketching method
(b) number line method
(c) table method
Solution
(a) (2x − 1)(x + 4)  x + 4
2x2 + 7x − 4  x + 4
2x2 + 6x − 8  0
x2 + 3x − 4  0
(x + 4)(x − 1)  0
y
y = (x + 4)(x – 1)
+
When (x + 4)(x − 1) = 0, x = − 4 or x = 1.
–4
The graph will intersect the x-axis at point x = – 4
and x = 1.
+
–
x
1
–4
Since (x + 4)(x − 1)  0, thus the range of values
of x is determined on the graph above the x-axis.
Hence, the range of values of x is x  − 4 or x  1.
(b)
Test point –5:
(–5 + 4)(–5 – 1)
Test point 0:
0 (0 + 4)(0 – 1)
+
x
–4 –4 –4
0
Test point 2:
(2 + 4)(2 – 1)
0
+
–
11
x
x
1
x
Since (x + 4)(x − 1)  0, then the range of values of x is determined on the positive part of
the number line.
Hence, the range of values of x is x  − 4 or x  1.
(c)
(x + 4)
(x − 1)
(x + 4)(x − 1)
x  −4
–
Range of values of x
−4  x  1
x1
–
+
–
+
+
–
+
+
Since (x + 4)(x – 1) > 0, then the range of values of x is determined on the positive part of
the table.
Hence, the range of values of x is x  − 4 or x  1.
2.1.3
43
CHAPTER 2
Self Practice 2.3
1. Solve each of the following quadratic inequalities by using graph sketching method, number
line method or table method.
(b) (2 – x)(8 – x) < 0
(c) x2  4x + 12
(a) x2 < 4
2
(f) (3x + 1)(5 – x) > 13
(d) x(x – 2)  3
(e) (x + 2) < 2x + 7
2. Find the range of values of x for 3x2 – 5x  16 + x(2x + 1).
Intensive Practice 2.1
Scan the QR code or visit bit.ly/2pSSNs6 for the quiz
1. Solve the quadratic equation 3x(x – 5) = 2x – 1. Give the answer in three decimal places.
2. Given a quadratic equation 2(x – 5)2 = 4(x + 7),
(a) express the equation in general form, that is ax2 + bx + c = 0.
(b) state the sum of roots and the product of roots of the equation.
3. If a and b are the roots of the quadratic equation 2x2 + 6x – 7 = 0, form equations with the
following roots:
5α 5β
1 , 1
(a)
(b)
(c) α + 3β, 3α + β
,
2α + 1 2β + 1
β α
4. If one of the roots of the equation 3x2 + 19x + k = 0 is –7, find the value of the constant k.
5. Given the quadratic equation rx2 + (r – 1)x + 2r + 3 = 0, where r is a non-zero integer, find the
value of r such that
(a) one root is negative of the other root,
(b) one root is the reciprocal of the other root,
(c) one root is twice the other root.
6. One root of the equation x2 – 8x + m = 0 is three times the other root, find the value of the
constant m and the roots.
7. The equation x2 + 2x = k(x – 1) has non-zero roots where the difference between the roots is
2, find the value of each root and the value of k.
8. The roots of the equation x2 + px + 27 = 0 are in the ratio of 1 : 3. Find the values of p.
9. Given 3 and h + 1 are the roots of the equation x2 + (k – 1)x + 9 = 0, find the possible
values of h and k.
10. The two roots of the equation x2 – 8x + c = 0 are a and a + 3d. Express c in terms of d.
11. Solve each of the following quadratic inequalities:
(a) 2x2 > x + 1
(b) (x – 3)2 < 5 – x
(c) (1 – x)2 + 2x , 17
12. Find the value of m and of n for each of the following quadratic inequalities:
(a) x2 + mx , n which is satisfied by –3 , x , 4.
(b) 2x2 + m . nx which is satisfied by x , –2 or x . 5.
13. Given y = 2x2 + bx + 12 and y , 0, if 2 , x , a, find the value of a and of b.
44
2.1.3
Quadratic Functions
2.2
Types of Roots of Quadratic Equations
Types of roots of quadratic equations and value of discriminant
InQuirY 3
In groups
CHAPTER 2
You have learned that the roots of a quadratic equation can be found by using the formula
–b ± b2 – 4ac
. Are the roots of a quadratic equation associated closely to the value of
x=
2a
b2 – 4ac in the formula? Let’s explore.
21st Century Learning
Aim: To explore the relation between types of roots of the
quadratic equation ax2 + bx + c = 0 and the value of b2 – 4ac
Instructions:
1. Scan the QR code or visit the link on the right.
bit.ly/2RS5Jff
2. Click one by one on the boxes which display the quadratic equations
y = x2 + 5x + 4, y = x2 – 6x + 9 and y = 9x2 – 6x + 2 to display each of the graphs.
3. Observe the positions of those graphs.
4. Identify the values of a, b and c as well as the roots of each graph when y = 0.
5. Discuss with the group members on the relation between the values of b2 – 4ac and
the types of roots obtained.
6. Present the findings of your group in front of the class.
From the results of Inquiry 3, note that the types of roots of
quadratic equations can be determined from the values of b2 – 4ac
which is known as the discriminant and usually denoted by D.
In general:
1. If the discriminant b2 – 4ac > 0, the equation has two
different real roots.
2. If the discriminant b2 – 4ac = 0, the equation has two
equal real roots.
3. If the discriminant b2 – 4ac < 0, the equation has no real
roots.
For the quadratic equation 9x2 – 6x + 2 = 0 which has no roots,
note that the value of the discriminant is negative. Since √ –36 is
not a real number, then this quadratic equation has no real roots.
The square root of a negative number is known as an imaginary
root and it is represented by i = √ –1. Then, the roots of the
quadratic equation 9x2 – 6x + 2 can be written as
6 ± √ –36(–1)
6 ± 6i
1±i
x=
=
=
.
18
18
3
2.2.1
Smart TIPS
When the discriminant
b2 – 4ac  0, the equation
has real roots.
Mind Challenge
What is the type of roots of
a quadratic equation if the
discriminant b2 – 4ac  0?
BRAINSTORMING
Determine the roots of
the following quadratic
equations. Give your
answers in terms of
imaginary number, i,
i = !w
–1.
(a) x2 + 4x + 5 = 0
(b) x2 – 2x + 3 = 0
(c) 2x2 – 6x + 5 = 0
45
CHAPTER 2
Example 6
Determine the type of roots for each of the following quadratic
equation.
(a) x2 + 5x – 6 = 0
(b) – 4x2 + 4x – 1 = 0
(c) 2x2 – 4x + 5 = 0
Mind Challenge
Why must the value of the
discriminant need to be
found first when determining
the types of roots of a
quadratic equation?
Solution
(a) x2 + 5x – 6 = 0 with a = 1, b = 5 and c = –6
b2 – 4ac = 52 – 4(1)(–6)
= 49 (> 0)
Thus, the equation x2 + 5x – 6 = 0 has two real and different
roots.
(b) – 4x2 + 4x – 1 = 0 with a = – 4, b = 4 and c = –1
b2 – 4ac = 42 – 4(– 4)(–1)
=0
Thus, the equation –4x2 + 4x – 1 = 0 has two equal real
roots.
(c) 2x2 – 4x + 5 = 0 with a = 2, b = – 4 and c = 5
b2 – 4ac = (– 4)2 – 4(2)(5)
= –24 (< 0)
Thus, the equation 2x2 – 4x + 5 = 0 has no real roots.
Tech Whizz
Check your answers by
Mathpapa application
which can be downloaded
from your mobile phone.
bit.ly/2LGCIgg
Self Practice 2.4
1. Find the discriminant and determine the types of roots for each of the following quadratic
equation:
(b) x2 = 8(x – 2)
(c) 5x2 + 4x + 6 = 0
(a) x2 + 4x + 1 = 0
(e) –x2 + 10x – 25 = 0
(f) (2x – 1)(x + 3) = 0
(d) –3x2 + 7x + 5 = 0
Solving problems involving types of roots of quadratic equations
The discriminant, D which determines the types of roots of a quadratic equation ax2 + bx + c = 0
can be used to:
(a) Find the value of a variable in the quadratic equation.
(b) Derive a relation.
Example 7
(a) The quadratic equation x2 + k + 3 = kx, where k is a constant, has two equal real roots.
Find the possible values of k.
(b) The roots of the equation (p + 2)x2 – 2px = 3 – p, where p is a constant, are real and
different. Find the range of values of p.
(c) Given the quadratic equation x2 + 4x + 13 = m(2 – x), where m is a constant, has no real
roots. Find the range of values of m.
46
2.2.1
2.2.2
Quadratic Functions
(a) x2 + k + 3 = kx
Arrange the equation in
x2 – kx + k + 3 = 0
the general form
a = 1, b = –k and c = k + 3
Two real and
b2 – 4ac = 0
equal roots
2
(–k) – 4(1)(k + 3) = 0
k2 – 4k – 12 = 0
(k + 2)(k – 6) = 0
k = –2 or k = 6
2
(b)
( p + 2)x – 2px = 3 – p
Arrange the equation in
( p + 2)x2 – 2px + p – 3 = 0
the general form
a = p + 2, b = –2p and c = p – 3
Two real and
b2 – 4ac > 0
different roots
(–2p)2 – 4( p + 2)( p – 3) > 0
4p2 – 4( p2 – p – 6) > 0
4p + 24 > 0
p > –6
2
(c)
x + 4x + 13 = m(2 – x)
x2 + 4x + 13 = 2m – mx
x2 + 4x + mx + 13 – 2m = 0
Arrange the equation in
x2 + (4 + m)x + 13 – 2m = 0
the general form
a = 1, b = 4 + m and c = 13 – 2m
No real roots
b2 – 4ac < 0
2
(4 + m) – 4(1)(13 – 2m) < 0
16 + 8m + m2 – 52 + 8m < 0
m2 + 16m – 36 < 0
(m + 18)(m – 2) < 0
+
+
Thus, the range of values of
m
–
–18
2
m is –18 < m < 2.
Example 8
Given the equation x2 – 4ax + 5b = 0 has two real and equal
roots, express a in terms of b.
Solution
x2 – 4ax + 5b = 0 where a = 1, b = – 4a and c = 5b.
Since the equation has two real and equal roots,
b2 – 4ac = 0
2
(– 4a) – 4(1)(5b) = 0
16a2 – 20b = 0
16a2 = 20b
a2 = 5 b
4
5b
a = ± 1 !w
2
2.2.2
BRAINSTORMING
By assuming b2 – 4ac  0,
show that the solutions of
the equation ax2 + bx + c = 0
are the reciprocal of the
solutions of the equation
cx2 + bx + a = 0.
MATHEMATICS
POCKET
Consider a line y = x + 4
that crossed a curve
y = x2 + 3x + m as shown in
the diagram below.
y
y=x+4
y = x2 + 3x + m
0
x
To find the range of values
of m, solve both equations
simultaneously.
x2 + 3x + m = x + 4
2
x + 2x + m – 4 = 0
This quadratic equation has
two real and different roots.
Thus,
b2 – 4ac > 0
22 – 4(1)(m – 4) > 0
4 – 4m + 16 > 0
4m < 20
m <5
Hence, the range of values of
m is m < 5. Discuss with your
friends and find the values
of m or the range of values
of m for the following cases:
• Line y = mx – 5 touches
a point on the curve
2y = x2 – 1.
• Line y = mx + 4 crosses a
curve 5x2 – xy = 2 at two
points.
• Line y = 2x + 3 does not
cross the curve x2 + xy = m.
47
CHAPTER 2
Solution
CHAPTER 2
Self Practice 2.5
1. Find the values or range of values of p such that the equation
(a) 9x2 + p + 1 = 4px has two equal roots,
(b) x2 + (2x + 3)x = p has two real and different roots,
(c) x2 + 2px + (p – 1)(p – 3) = 0 has no real roots.
2. Find the range of values of k if the equation x2 + k = kx – 3 has two real and different roots.
State the values of k if the equation has two real and equal roots.
3. The quadratic equation x2 + hx + k = 0 has roots of –2 and 6, find
(a) the value of h and of k,
(b) the range of values of c such that the equation x2 + hx + k = c has no real roots.
4. The equation hx2 + 3hx + h + k = 0, where h ≠ 0, has two real and equal roots. Express k in
terms of h.
5. Given the quadratic equation ax2 – 5bx + 4a = 0, where a and b are constants, has two real
and equal roots, find a : b.
Intensive Practice 2.2
Scan the QR code or visit bit.ly/2nCIGqQ for the quiz
1. Determine the types of roots for the following quadratic equations.
(a) x2 – 8x + 16 = 0
(b) (x – 2)2 = 3
(c) 2x2 + x + 4 = 0
2. The following quadratic equations have two real and equal roots. Find the values of k.
(a) x2 + kx = 2x – 9
(b) kx2 + (2k + 1)x + k – 1 = 0
3. The following quadratic equations have two real and different roots. Find the range of
values of r.
(a) x(x + 1) = rx – 4
(b) x2 + x = 2rx – r2
4. Find the range of values of p if the following equations have no real roots.
(a) (1 – p)x2 + 5 = 2x
(b) 4px2 + (4p + 1)x + p – 1 = 0
5. The equation kx2 – 10x + 6k = 5, where k is a constant, has two real and equal roots.
(a) Find the values of k.
(b) Hence, find the roots of the equation by using the smallest value of k obtained in (a).
6. The quadratic equation x(x – 4) + 2n = m where m and n are constants, has two real and
equal roots. Express m in terms of n.
7. The quadratic equation x2 + bx + c = 0 where b and c are positive integers, has a
discriminant of 16 and b – c = –4. Find
(a) the possible values of b and c,
(b) the corresponding roots of the equations.
8. The quadratic equation 2x2 – 5x + c = 0 where c is a positive integer, has no real roots.
(a) Find two possible values of c, that is c1 and c2.
1
(b) Based on the values of c1 and c2 in (a), does the equation 2x2 – 5x + (c1 + c2) = 0 have
2
two real roots? Explain.
48
2.2.2
Quadratic Functions
2.3
Quadratic Functions
CHAPTER 2
A ball is thrown into the net. What can you observe from the
path of the ball? If you observe the path of the ball, it follows
the shape of a parabola. The path or the curve is the shape of
the graph of a quadratic function. What are other examples that
involve the shape of a parabola?
Analysing the effect of changes of a, b and c towards the shape and
position of the graph for f (x) = ax2 + bx + c
The general form of a quadratic function is a function in the form of f(x) = ax2 + bx + c where
a, b, and c are constants and a ≠ 0. The shape of the graph of a quadratic function is a parabola
which is symmetrical about the axis that passes through the minimum point or maximum point.
f(x)
Axis of symmetry
y = f(x)
f(x)
0
x
y = f(x)
x
0
Maximum point
MATHEMATICS
POCKET
The highest power of a
quadratic function is the
same as the highest power
of a quadratic equation,
that is 2.
Axis of symmetry
Minimum point
What are the effects on the shape and position of the quadratic function graph if the values
of a, b and c change? Let’s explore.
InQuirY 4
In groups
21st Century Learning
Aim: To explore the effects of changes in the values of a, b and c
towards the shape and position of the quadratic function graph
Instructions:
1. Scan the QR code or visit the link on the right.
bit.ly/324HT0w
2. Observe the graph of function f(x) = ax2 + bx + c, where a = 1,
b = 2 and c = 3.
3. In groups, analyse the changes of shape and position of the quadratic function graph
based on the following instructions:
(a) Drag slide a to the left and to the right without changing slide b and slide c.
(b) Drag slide b to the left and to the right without changing slide a and slide c.
(c) Drag slide c to the left and to the right without changing slide a and slide b.
4. Make a generalisation on the effects of changes in the values of a, b and c on the
shape and position of the graph of f(x) = ax2 + bx + c.
5. Present the findings of your group in front of the class and discuss with other groups.
2.3.1
49
From the results of Inquiry 4, the following findings are obtained.
•
CHAPTER 2
•
Only the
•
value of a
changes
•
•
Only the •
value of b •
changes
•
Only the •
value of c
changes •
Changes in shape and position of the graph of function f(x) = ax2 + bx + c
Change in value of a affects the shape and width of the graph, however the
y-intercept remains unchanged.
When a . 0, the shape of the graph is which passes through the minimum point and
when a , 0, the shape of the graph is which passes through the maximum point.
For the graph a . 0, for example a = 1, when the value of a is larger than 1, the
width of the graph decreases. Conversely, when the value of a is smaller than 1 and
approaches 0, the width of the graph increases.
For the graph a , 0, for example a = –1, when the value of a is smaller than –1, the
width of the graph decreases. Conversely, when the value of a increases from ­–1 and
approaches 0, the width of the graph increases.
Change in value of b only affects the position of vertex with respect to the y-axis,
however the shape of the graph and the y-intercept are unchanged.
When b = 0, the vertex is on the y-axis.
For the graph a . 0, when b . 0, the vertex is on the left side of the y-axis and
when b , 0, the vertex is on the right side of the y-axis.
For the graph a , 0, when b . 0, the vertex is on the right side of the y-axis and
when b , 0, the vertex is on the left side of the y-axis.
Change in value of c only affects the position of graph either vertically upwards or
vertically downwards.
The shape of the graph is unchanged.
Example 9
The diagram shows the sketch of the graph for f(x) = x2 + 1 where
a = 1, b = 0 and c = 1. Make an analysis and a generalisation on
the shape and position of the graph when the following values change.
Hence, sketch the graph.
(a) The value of a becomes
(i) 2, (ii) 1 .
2
(b) The value of c becomes 3.
Solution
(a) (i) When a changes from 1 to 2, the
width of the graph decreases. The
y-intercept does not change and
the vertex is on the y-axis.
f(x)
50
f(x) = x2 + 1
1
x
0
1
,
2
the width of graph increases. The
y-intercept does not change and the
vertex is on the y-axis.
(ii) When a changes from 1 to
f(x)
f(x) = x2 + 1
f(x) = 2x2 + 1
f(x)
f(x) = x2 + 1
1
0
x
1
0
1
f(x) = –– x2 + 1
2
x
2.3.1
Quadratic Functions
(b) When c changes from 1 to 3, the shape of the graph remains
unchanged. The only change is the position, that is, the graph
moves 2 units upwards.
f(x)
3
f(x) = x2 + 1
1
0
x
Self Practice 2.6
f(x)
1. The diagram on the right shows the graph for f(x) = –x2 + x + 6,
where a = –1, b = 1 and c = 6. Sketch the graph of f(x) formed
6
when the following values change.
f(x) = –x2 + x + 6
(a) The value of a changes to
(i) –3 (ii) – 1 ,
4
(b) The value of b changes to –1,
x
–2 0
3
(c) The value of c changes to –2.
Make a generalisation from the changes in the shape and position of the graphs obtained.
Relating the position of the graph of a quadratic function and the
types of roots
You have learned that the discriminant b2 – 4ac of a quadratic equation ax2 + bx + c = 0
can be used to determine the types of roots. Let’s see how the types of roots of a quadratic
equation can determine the position of the graph of a quadratic function f(x) = ax2 + bx + c with
respect to the x-axis.
InQuirY 5
In groups
Aim: To explore the relation between the position of the graph of quadratic function and
types of roots
Instructions:
1. Each group has to choose only one case out of the following two cases.
Case 1
(a) f(x) = x + 4x + 4
(b) f(x) = 2x2 + 7x – 4
(c) f(x) = x2 – 6x + 12
2
Case 2
(a) f (x) = –x2 + 2x – 1
(b) f (x) = –2x2 – 8x – 5
(c) f (x) = –x2 + 6x – 10
2. By using a dynamic geometry software, plot the graph of each quadratic function in the
case chosen.
3. Observe the shapes of the graphs obtained and the respective roots.
4. State the relation between the value of b2 – 4ac, types of roots and the number of
intersection points on the x-axis.
5. From the relation, state the position of the graph of the quadratic function obtained.
6. Compare the findings of your group with other groups of different case and make a
conclusion on the comparison made.
2.3.1
2.3.2
51
CHAPTER 2
f(x) = x2 + 3
CHAPTER 2
From the results of Inquiry 5, the relation between the position of graph of quadratic function
f(x) = ax2 + bx + c on the x-axis and the types of roots can be summarised as shown in the table
below.
Discriminant,
b2 – 4ac
Types of roots and
position of graph
b2 – 4ac > 0
• Two real and
different roots
• The graph intersects
the x-axis at two
different points.
b2 – 4ac = 0
• Two real and equal
roots
• The graph touches
the x-axis at one
point only.
b2 – 4ac < 0
• No real roots
• The graph does not
intersect at any point
on the x-axis.
Position of graph of function f(x) = ax2 + bx + c
a>0
α
β
a<0
x
α
β
α=β
α =β
x
x
x
x
x
Example 10
Determine the types of roots for each of the following quadratic functions when f(x) = 0. Then,
sketch the graph and make a generalisation on the position of graph at the x-axis.
(b) f(x) = –x2 + 2x – 1
(a) f(x) = 2x2 + x – 5
Solution
(a) f(x) = 2x2 + x – 5
a = 2, b = 1, c = –5
b2 – 4ac = (1)2 – 4(2)(–5)
= 41 (> 0)
The quadratic function has two real and different roots. Since
a > 0, thus the graph of f(x) is a parabola which passes through
the minimum point and intersects the x-axis at two points.
(b) f(x) = –x2 + 2x – 1
a = –1, b = 2, c = –1
b2 – 4ac = (2)2 – 4(–1)(–1)
=0
The quadratic function has two real and equal roots. Since
a < 0, thus the graph of f(x) is a parabola which passes through
the maximum point and intersects the x-axis at one point.
52
x
x
2.3.2
Quadratic Functions
(a) Find the values of m such that the x-axis is the tangent to the graph of a quadratic
function f (x) = (m + 1)x2 + 4(m – 2)x + 2m.
(b) Find the range of values of k if the graph of a quadratic function f (x) = 2x2 + 5x + 3 – k
has no x-intercept.
(c) Find the range of values of p if the graph of a quadratic function f (x) = x2 + px + p + 3
has two x-intercepts.
Solution
(a) The graph of a quadratic function f (x) = (m + 1)x2 + 4(m – 2)x + 2m such that the x-axis is
a tangent to the graph which means the function has two real and equal roots.
For two real and equal roots:
b2 – 4ac = 0
(4m – 8)2 – 4(m + 1)(2m) = 0
16m2 – 64m + 64 – 8m2 – 8m = 0
8m2 – 72m + 64 = 0
m2 – 9m + 8 = 0
(m – 1)(m – 8) = 0
m = 1 or m = 8
(b) The graph of a quadratic function f (x) = 2x2 + 5x + 3 – k has no x-intercept, which means
the function has no real roots.
For no real roots:
b2 – 4ac < 0
52 – 4(2)(3 – k) < 0
25 – 24 + 8k < 0
1 + 8k < 0
8k < –1
k<– 1
8
(c) The graph of a quadratic function f (x) = x2 + px + p + 3
has two x-intercepts which means the function has two
real and different roots.
For two real and different roots:
b2 – 4ac > 0
2
p – 4(1)( p + 3) > 0
p2 – 4p – 12 > 0
(p + 2)(p – 6) > 0
p < – 2 or p > 6
+
What are the rules of
a quadratic function
f(x) = ax2 + bx + c so that it
is always positive or always
negative for all real values
of x? Discuss.
+
–2
2.3.2
BRAINSTORMING
–
6
p
53
CHAPTER 2
Example 11
CHAPTER 2
Self Practice 2.7
1. Determine the types of roots for each of the following quadratic function. Sketch the graph
and make generalisation on the position of the graph on the x-axis.
(b) f (x) = x2 + 2x – 3
(c) f (x) = 4x2 – 8x + 5
(a) f (x) = –3x2 + 6x – 3
2. Find the possible values of h if the graphs of the following quadratic functions touch the
x-axis at only one point.
(b) f (x) = x2 – (h + 3)x + 3h + 1
(a) f (x) = x2 – 2hx + 2 + h
3. Find the range of values of q if the graph of the following quadratic functions intersect the
x-axis at two points.
(b) f (x) = (q + 2)x2 + q(1 – 2x) – 5
(a) f (x) = 5x2 – (qx + 4)x – 2
4. Find the range of values of r if the graphs of the following quadratic functions do not
intersect the x­-axis.
(b) f (x) = rx2 + (2r + 4)x + r + 7
(a) f (x) = rx2 + 4x – 6
Making relation between the vertex form of a quadratic function
f (x) = a (x – h)2 + k with the other forms of quadratic functions
The diagram on the right shows the sketch of graph of a
quadratic function in the vertex form, f (x) = (x – 2)2 – 9. Since
a > 0, the graph of the quadratic function is in the shape of .
Note that the graph of this quadratic function has its vertex
at the minimum point (2, –9) and the equation of the axis of
symmetry, x = 2.
The vertex form is a quadratic function in the form of
f(x) = a(x – h)2 + k, where a, h and k are constants. The vertex
is (h, k) and it is symmetrical about the line x = h.
When a > 0, the vertex (h, k) is the minimum point and k
is the minimum value of f (x). When a < 0, the vertex (h, k) is
the maximum point and k is the maximum value of f(x).
f(x)
f(x) = (x – 2)2 – 9
–1 0
x
5
–5
(2, –9)
x=2
Other than the vertex form, the quadratic function can be written in the following form:
Form of quadratic functions
• General form, f (x) = ax2 + bx + c, where
a, b and c are constants with a vertex at
(
( ))
the point – b , f – b and symmetrical
2a
2a
about the line x = – b .
2a
• Intercept form, f (x) = a(x – p)(x – q),
where a, p and q are constants. p and q are
the roots or x-intercepts for f (x), its vertex is
p+q
p+q
at the point
and
,f
2
2
p+q
symmetrical about the line x =
.
2
(
(
))
What is the relation that exists between vertex form of quadratic functions with the general
form and the intercept form? Let’s explore.
54
2.3.2
2.3.3
Quadratic Functions
In groups
Aim: To explore the relation between the vertex form of a quadratic function with the
general form and intercept form
Instructions:
1. Consider a quadratic function in the vertex form, f (x) = (x – 4)2 – 4.
2. In each group, discuss and express the quadratic function in the vertex form into
general form and intercept form.
3. Then, copy and complete the table below.
Shape of
quadratic
function
Quadratic
function
Vertex form
f (x) = (x – 4)2 – 4
x-intercept y-intercept Vertex
Axis of
symmetry
General form
Intercept form
4. Sketch the graphs for each of the forms of the quadratic function. Check the sketching
of graphs by using dynamic geometry software.
5. Compare the graphs of quadratic function which are plotted in the vertex form,
general form and intercept form.
6. Carry out a brainstorming session in the group and obtain a conclusion on the relation
that exists between quadratic function in the vertex form with the general form and
intercept form.
From the results of Inquiry 6, it is found that the quadratic function f (x) = (x – 4)2 – 4 in vertex
form, general form and intercept form produced the same graph when they are sketched.
In expressing quadratic function in the vertex form to general form and intercept form or
vice versa, the following methods can be used:
Expansion
Factorisation or formula
f (x) = a(x – h)2 + k
f (x) = ax2 + bx + c
f (x) = a(x – p)(x – q)
Vertex forms
General form
Intercept form
Completing the square
2.3.3
Expansion
55
CHAPTER 2
InQuirY 6
Example 12
(
CHAPTER 2
)
9
1
–
in the
4
8
intercept form, f(x) = a(x – p)(x – q), where a, p and q are
constants and p , q. Hence, state the values of a, p and q.
Express quadratic function, f(x) = 2 x +
2
Solution
Convert the vertex form of the quadratic function into the
general form first.
2
f(x) = 2 x + 9 – 1
4
8
9
= 2 x2 + x + 81 – 1
2
16 8
General form
= 2x2 + 9x + 10
= (2x + 5)(x + 2)
Intercept form
= 2 x + 5 (x + 2)
2
Thus, the quadratic function in the intercept form for
2
f(x) = 2 x + 9 – 1 can be expressed as
4
8
5
f(x) = 2 x + (x + 2), where a = 2, p = – 5 and q = –2.
2
2
(
(
)
(
(
(
)
)
)
)
Mind Challenge
Not all the vertex forms
or general forms can be
expressed in intercept form,
only graphs which have
x-intercept can be expressed.
Do you agree with the
statement?
Alternative Method
( )
[( ) ]
2
f(x) = 2 x + 9 – 1
4
8
2
1
9
– 2
=2 x+
4
4
Use a2 – b2 = (a + b)(a – b)
f(x) = 2 x + 9 + 1 x + 9 – 1
4 4
4 4
10
8
=2x+
x+
4
4
5
= 2 x + (x + 2)
2
(
(
(
)(
)
)(
)
)
Example 13
Express f(x) = –3x2 + 2x + 1 as f(x) = a(x – h)2 + k, where a,
h and k are constants. Hence, determine the values of a, h
and k.
Solution
f(x) = –3x2 + 2x + 1
Make sure the coefficient of x2 is 1 before completing the square.
f(x) = –3x2 + 2x + 1
Factorise –3 from –3x2 + 2x + 1
= –3 x2 – 2 x – 1
3
3
2
2
Add and subtract
= –3 x2 – 2 x + –1 – –1 – 1
coefficient of x 2
3
3
3
3
2
2
2
= –3 x – 1 – –1 – 1
3
3
3
2
1
4
= –3 x –
–
3
9
2
1
4
= –3 x –
+
3
3
1
Thus, a = –3, h = and k = 4 .
3
3
(
[
[( )
[( )
( )
56
)
( ) ( ) ]
( ) ]
]


BRAINSTORMING
By using completing the
square method, show
that the equation of the
axis of symmetry for
f(x) = ax2 + bx + c is
x=– b .
2a
2.3.3
Quadratic Functions
Self Practice 2.8
2. Express each of the following vertex form into general form and intercept form.
(b) f(x) = 9 – (2x – 1)2
(c) f(x) = 2(x + 1)2 – 18
(a) f(x) = (x – 2)2 – 1
3. Find the vertex of the function f(x) = – 1 (x + 4)2 – 5 and convert it into general form.
2
f(x)
4. The diagram on the right shows the graph of quadratic
2
function f(x) = a(x + h) + k, where a, h and k are constants.
Given (–2, 16) is the maximum point of the graph.
(–2, 16)
(a) State the values of a, h and k.
12
(b) Hence, express the function in general form,
2
f(x) = ax + bx + c and intercept form,
f(x) = a(x + h)2 + k
f(x) = a(x – p)(x – q).
0
x
5. Express each of the following in the vertex form, f(x) = a(x + h)2 + k, where a, h and k are
constants.
(b) f(x) = –x2 – 2x + 4
(c) f(x) = –2x2 – x + 6
(a) f(x) = x2 – x – 6
2
(e) f(x) = (x + 2)(6 – x)
(f) f(x) = 2(x + 4)(x – 2)
(d) f(x) = 3x – 2x – 9
Analysing the effect of change of a, h and k on the shape and position of graph for f (x) = a(x – h)2 + k
The quadratic function in the vertex form, f(x) = a(x + h)2 + k, where a, h and k are constants
has its vertex at (h, k) and it is symmetrical about the line x = h. What will happen to the
shape and position of the graph of function f(x) when the values of a, h and k are changed?
InQuirY 7
In groups
21st Century Learning
Aim: To explore the effect of change in the values of a, h and k
on the shape and position of the graph f(x) = a(x – h)2 + k
Instructions:
1. Scan the QR code or visit the link on the right.
bit.ly/2OrShvq
2. Observe the graph of function f(x) = a(x – h)2 + k where a = 2,
h = 3 and k = 1.
3. In groups, make an analysis and state the observation on the shape and position of the
graph based on each of the following instructions:
(a) Drag slide a to the left and to the right without changing slide h and slide k.
(b) Drag slide h to the left and to the right without changing slide a and slide k.
(c) Drag slide k to the left and to the right without changing slide a and slide h.
4. What happens to the axis of symmetry, minimum value or maximum value of the graph
when the values of a, h or k change?
5. Make a generalisation on the effect of change in the values of a, h and k on the shape
and position of the graph f(x) = a(x – h)2 + k.
2.3.3
2.3.4
57
CHAPTER 2
1. Given f(x) = 2(x – 3)2 – 8 = a(x – p)(x – q) for all values of x, find the values of the constants
a, p and q where p , q.
From the results of Inquiry 7, it is found that:
CHAPTER 2
•
•
Only the
value of a
changes
Only the
value
of h
changes
•
•
•
•
•
•
•
Only the •
value of k
changes •
Changes in shape and position of the graph of function f(x) = a(x – h)2 + k
Change in value of a affects the shape and width of graph.
When a . 0, the graph is in the shape of which passes through the minimum point and
when a , 0, the graph is in the shape of which passes through the maximum point.
For the graph a . 0, for example a = 2, when the value of a is increased to more
than 2, the width of the graph decreases. Conversely, when the value of a decreases
from 2 and approaches 0, the width of the graph increases.
For the graph a , 0, for example a = –2, when the value of a is decreased to smaller
than –2, the width of the graph decreases. Conversely, when the value of a increases
from –2 and approaches 0, the width of the graph increases.
The axis of symmetry and the maximum or minimum value remain unchanged.
The change in the values of h only shows the horizontal movement of the graph.
When the value of h increases, the graph will move to the right whereas when the
value of h decreases, the graph will move to the left.
The position of the axis of symmetry changes but the minimum or maximum values
remain unchanged.
The change in the values of k only shows vertical movement of the graph.
When the values of k increases, the graph will move upwards whereas when the
value of k decreases, the graph will move downwards.
The maximum and minimum values change but the axis of symmetry remain
unchanged.
Example 14
f(x)
The diagram on the right shows the graph of f(x) = 2(x + 2)2 + 3, where
a = 2, h = –2 and k = 3. Make generalisation on the effect of change in
11
each of the following values on the shape and position of the graph.
(a) The value of a changes to
f(x) = 2(x + 2)2 + 3
(i) 6, (ii) 1 .
2
3
(b) The value of h changes to –6.
–2 0
(c) The value of k changes to 8.
Solution
(a) (i) When a changes from 2 to 6, the
width of the graph decreases. The
axis of symmetry and the minimum
value remain unchanged.
f(x)
f(x) = 2(x + 2)2 + 3
58
f(x)
11
f(x) = 2(x + 2)2 + 3
3
–2 0
1
, the
2
width of the graph increases. The axis
of symmetry and the minimum value
remain unchanged.
(ii) When a changes from 2 to
1
f(x) = – (x + 2)2 + 3
2
11
f(x) = 6(x + 2)2 + 3
x
x
3
–2 0
x
2.3.4
Quadratic Functions
(c) When k changes from 3 to 8, the graph
with the same shape moves vertically 5
units upwards. The minimum value
becomes 8 and the equation of the axis of
symmetry is still the same, that is x = –2.
CHAPTER 2
(b) When h changes from –2 to – 6, the graph
with the same shape moves horizontally
4 units to the left. The equation of the
axis of symmetry becomes x = – 6 and the
minimum value remains unchanged, that
is 3.
f(x)
f(x)
f(x) = 2(x + 2)2 + 3
11
16
f(x) = 2(x + 2)2 + 8
8
3
f(x) = 2(x + 6)2 + 3
–6
11
–2 0
x
f(x) = 2(x + 2)2 + 3
3
x
–2 0
Self Practice 2.9
1. The diagram on the right shows the graph of f(x) = –3(x – 2)2 + 4
where a = –3, h = 2 and k = 4.
(a) Determine the coordinate of maximum point and the equation
of the axis of symmetry.
(b) Make generalisation on the shape and position of the graph
when the following values change. Hence, sketch the graphs.
(i) The value of a changes to –10.
(ii) The value of h changes to 5.
(iii) The value of k changes to –2.
2. The diagram on the right shows the graph of function
f(x) = (x – 3)2 + 2k, where k is a constant. Given (h, –6) is the
minimum point of the graph.
(a) State the values of h, k and p.
(b) If the graph moves 2 units to the right, determine the
equation of the axis of symmetry for the curve.
(c) If the graph moves 5 units upwards, determine the
minimum value.
f(x)
f(x) = –3(x – 2)2 + 4
4
0
2
f(x)
x
f(x) = (x – 3)2 + 2k
p
x
0
(h, – 6)
3. Compare the graph of each of the following quadratic functions to the graph of f(x) = x2 with
its vertex at (0, 0).
(a) f(x) = 1 (x – 6)2
(b) f(x) = 3(x – 1)2 + 5
(c) f(x) = 1 (x + 1)2 – 4
2
4
2.3.4
59
Sketching the graphs of quadratic functions
CHAPTER 2
The graph of a quadratic function in various shapes can be sketched by following the steps below:
Identify the value of
a to determine the
shape of the graph of a
quadratic function.
Find the value of
discriminant, b2 – 4ac
to determine the
position of the graph.
Plot the points obtained
on the Cartesian
plane and draw a
smooth parabola that
is symmetrical at the
vertical line passing
through the vertex.
Find the value of
f(0) to determine the
y-intercept.
Sketch the graph of quadratic function f(x) = –x2 + 4x + 12.
Solution
a < 0, so f(x) has a maximum point
b2 – 4ac = 42 – 4(–1)(12)
= 16 + 48
= 64 (> 0)
The curve intersects the x-axis at two different points.
f(x) = –x2 + 4x + 12
= – (x2 – 4x – 12)
2
2
= – x2 – 4x + – 4 – – 4 – 12
2
2
= – (x – 2)2 + 16
Maximum point is (2, 16) and the equation of the axis of
symmetry, x = 2.
f(x) = 0
–x2 + 4x + 12 = 0
(–x + 6)(x + 2) = 0
–x + 6 = 0 or x + 2 = 0
x =6
x = –2
The intersection on the x-axis are x = –2 and x = 6.
f (0) = – (0)2 + 4(0) + 12
= 12
The graph intersects the y-axis at (0, 12).
60
Determine the
intersection point on the
x-axis by solving the
equation of quadratic
function f(x) = 0.
BRAINSTORMING
Example 15
[
Determine the vertex.
( ) ( )
How to sketch a quadratic
function graph in
Example 15 in the domain
–3  x  7?
]
BRAINSTORMING
Without expressing to
vertex form, how to find
the vertex of a quadratic
function in general
form, f(x) = ax2 + bx + c
and intercept form,
f(x) = a(x – p)(x – q)? Discuss.
2.3.5
Quadratic Functions
f(x)
12
–2 0
x=2
(2, 16)
f(x) = –x2 + 4x + 12
6
CHAPTER 2
The curve is sketched as shown in the
diagram on the right.
x
Self Practice 2.10
1. Sketch the graphs of the following quadratic functions.
(b) f(x) = 2(x + 2)2 – 2
(a) f(x) = (x – 1)2 – 4
(d) f(x) = –2(x – 1)(x – 3)
(e) f(x) = –(x + 3)(x + 5)
(h) f(x) = 2x2 + 3x – 2
(g) f(x) = –x2 + 4x + 5
(c) f(x) = 9 – (x – 2)2
(f) f(x) = 2(x + 1)(x – 3)
(i) f(x) = –x2 + 4x + 12
Solving problems of quadratic functions
The knowledge on quadratic function is very important and it is widely used in our daily lives.
The graph of a quadratic function in the shape of a parabola helps us to solve many problems.
For example, it can be used to forecast the profit or loss in business, plot the curved motion of
an object and determine the minimum or maximum values.
Example 16
Mathematics
Application
Suresh is chosen to represent his school in the district level javelin competition. He throws
the javelin at a distance of 3 metres from the ground. The height of the javelin is given by the
function h(t) = –5t2 + 14t + 3, where h is the height, in metres, of the javelin and t is the time,
in seconds.
(a) Find the maximum height, in metres, of the javelin thrown by Suresh.
(b) Calculate the time, in seconds, when the javelin touches the ground.
Solution
1 . Understanding the problems
The function of the height of javelin is h(t) = –5t2 + 14t + 3, where h is the height of
javelin, in metres, and t is the time, in seconds, after the javelin is thrown.
2 . Planning a strategy
◆
◆
2.3.5
Express the quadratic function in vertex form and determine the maximum value.
Solve the equation h(t) = 0 to find the intercept on the t-axis, that is the time for the
javelin to touch the ground.
2.3.6
61
CHAPTER 2
3 . Implementing the strategy
(a) h(t) = –5t2 + 14t + 3
Make the coefficient of t2 as 1
= –5 t2 – 14 t – 3
5
5
coefficient of t
Add and subtract
2
2
14
2
2
= –5 t –
t+ – 7 – – 7 – 3
5
5
5
5
2
= –5 t – 7 – 64
5
25
7 64
2
Vertex is ,
7
64
5 5
= –5 t –
+
5
5
Since a < 0, then the maximum value of h(t) is 64 when t = 7 .
5
5
64
metres = 12.8 metres.
Thus, the maximum height reached by the javelin is
5
(b)
h(t) = 0
–5t + 14t + 3 = 0
5t2 – 14t – 3 = 0
(5t + 1)(t – 3) = 0
t = – 1 (rejected) or t = 3
5
Thus, the time when the javelin touches the ground is 3 seconds.
(
)
(
( ) ( ) )
[( ) ]

( )


2

4 . Making a conclusion
Function h(t) = –5t2 + 14t + 3.
(a) Coordinates of the maximum height:
t=– b
2a
= – 14
2(–5)
= 1.4
Substitute t = 1.4 into the quadratic function,
h(1.4) = –5(1.4)2 + 14(1.4) + 3
= 12.8
Thus, the maximum height achieved by the javelin is 12.8 metres after 1.4 seconds.
(b) When the time is 3 seconds:
h(t) = –5(3)2 + 14(3) + 3
= – 45 + 42 + 3
= 0
62
2.1.1
2.3.6
Quadratic Functions
1. The function h(t) = –5t2 + 8t + 4 represents the height h, in metres, of a diver from the water
surface in a swimming pool, t seconds after he dives from the diving board. Find
(a) the height, in metres, of the diving board from the water surface,
(b) the time, in seconds, the diver achieves at the maximum height,
(c) the maximum height, in metres, the diver achieves,
(d) the time, in seconds, when the diver is in the air.
2. A tunnel at a certain part of a highway is in the shape of
parabola. The height, in metres, of the curve of the parabola is
given by the function h(x) = 15 – 0.06x2, where x is the width of
the tunnel, in metres.
(a) Determine the maximum height, in metres, of the tunnel.
(b) Find the width, in metres, of the tunnel.
3. The diagram on the right shows the cross section of
a parabolic satellite whose function is represented by
1
f(x) = x2, where x and y are measured in metres. Find
4
the width and the depth of the parabola, in metres.
4. The diagram on the right shows a suspension bridge.
The function of the cables between the two pillars of the
1 2
bridge is represented by y =
x – x + 150, where x
400
and y are measured in metres. The minimum point of
the cable is on the road surface at the middle of the two
pillars.
(a) What is the distance between the minimum point
and each pillar?
(b) What is the height of the road above the water level?
Intensive Practice 2.3
y
y = 1– x2
4
(–2, 1)
0
(2, 1)
x
y
1
y = ––– x2 – x + 150
400
0
Road
x
Pillars
Scan the QR code or visit bit.ly/33jjtRj for the quiz
1. Find the values or range of values of k, if the quadratic function
(a) f(x) = kx2 – 4x + k – 3 has only one intercept,
(b) f(x) = 3x2 – 4x – 2(2k + 4) intersects the x-axis at two different points.
2. Find the smallest value of integer m such that the function f(x) = mx2 + 7x + 3 is always
positive for all real values of x.
3. The quadratic function is defined by f(x) = x2 + 6x + n, where n is a constant.
(a) Express f(x) in the form (x – h)2 + k, where h and k are constants.
(b) Given the minimum value of f(x) is –5, find the value of n.
(c) Sketch the curve of f(x).
2.3.6
63
CHAPTER 2
Self Practice 2.11
CHAPTER 2
4. Find the range of values of r such that the line y = rx + 4 does not intersect the curve
y = x2 – 4x + 5. State the values of r such that the line y = rx + 4 is a tangent to the curve
y = x2 – 4x + 5.
5. Explain the effect on the shape and position of the graph for each change on the following
quadratic functions.
(a) Change f(x) = 3(x – 1)2 + 2 to f(x) = 6(x – 1)2 + 2.
(b) Change f(x) = 3(x – 1)2 + 2 to f(x) = 3(x – 4)2 + 2.
(c) Change f(x) = 3(x – 1)2 + 2 to f(x) = 3(x – 1)2 + 5.
6. The height, h, in metres, of a bird to catch a fish in a pond can
be represented by the function h(t) = 2(t – 3)2, where t is the
time, in seconds, when the bird starts to fly to catch the fish.
(a) Sketch the graph of h(t).
(b) The movement of another bird is represented by the function
r(t) = 2h(t). Sketch the graph of r(t).
(c) Compare the graphs of h(t) and r(t). Which bird starts
to move at the highest position? Explain.
7. Given a quadratic function f(x) = 3 – 4k – (k + 3)x – x2, where k is a constant, is always
negative when p , k , q. Find the value of p and of q.
f(x)
8. The diagram on the right shows a bridge PQ of length 8 m
across a river. The supporting cable ABC on the bridge can be
1
represented by the function f(x) = x2 + bx + c, where b and
8
c are constants.
(a) Find the value of b.
(b) Find the value of c such that the minimum point B on
the cable is always above PQ.
(c) Find the values of c if b is 2 m above PQ.
9. The function h(t) = –4t 2 + 32t as shown in the diagram on
the right represents the height, in metres, of a firework,
t seconds after it was launched. The fireworks exploded at
the highest point.
(a) When did the firework explode?
(b) What was the height at which the fireworks explode?
10. The diagram on the right shows the graph of y = –(x – a)(x – b)
where a , b.
(a) Given that M is the midpoint of AB, express the
following lengths, in terms of a and/or b.
(i) OA
(ii) OB
(iii) OC
(iv) OM
a+b
and – ab in the diagram
(b) Can you interpret
2
geometrically?
1 x2 + bx + c C
A y = ––
8
B
8m
0 P
Q
h(t) = – 4t 2 + 32t
y
y = – (x – α)(x – β )
0
A
B
x
C
11. The maximum value of f(x) = x2 – 4nx + 5n2 + 1 is m2 + 2n where m and n are constants.
Show that m = n – 1.
64
x
Quadratic Functions
me
th
o
ng
Quadratic
inequalities
o
olv
fs
g
in
Graph sketching
Number line
Table
a > 0, graph in the shape of
a < 0, graph in the shape of
Quadratic functions
hanges
QUADRATIC
FUNCTIONS
c
ec
eff
For a > 0,
b > 0, graph shifts to the left
b < 0, graph shifts to the right
ts
Discriminant
D = b2 – 4ac
Type of roots
od
me
th
ing
so l v
Quadratic equation
ax2 + bx + c = 0
involves
(x – )(x – ) = 0
or
=6
x2 – ( + )x +
of
of
Formula
on
fo
equ rmi
at
i
d
Completing the
square
CHAPTER 2
Summary OF Chapter 2
Two equal roots
D=0
No real roots
D<0
c increases, graph shifts upwards
c decreases, graph shifts downwards
of changes
cts
e
f
f
a > 0, graph in the shape of
a < 0, graph in the shape of
Vertex form
2
f(x) = a(x – h) + k
e
Two real and
distinct roots
D>0
General form
f(x) = ax2 + bx + c
Intercept form
f(x) = a(x – p)(x – q)
h increases, graph shifts to the right
h decreases, graph shifts to the left
k increases, graph shifts upwards
k decreases, graph shifts downwards
WRITE YOUR JOURNAL
The word quadratic originates from the word quad which means four but a quadratic
equation involves polynomial with the maximum power of 2. Carry out a study on the
origin of the word quadratic that is related to quadratic equation. Produce a graphical folio
on your study.
65
MASTERY PRACTICE
CHAPTER 2
1. Solve the quadratic equation 3x(x – 4) = (2 – x)(x + 5). Write the answer in three decimal
places. PL2
2. Given the quadratic equation (x – 4)2 = 3. PL2
(a) Express the equation in the general form, ax2 + bx + c = 0.
(b) State the sum of roots and products of roots of the equation.
(c) Determine the type of roots of the equation.
3. Find the values of k or the range of values of k such that the equation x2 + kx = k – 8 has
(a) two equal roots,
(b) two real and different roots,
(c) real roots.
PL2
4. Given the quadratic equation 3x2 + px – 8 = 0, where p is a constant. Find the value of p if
PL2
(a) one of the roots of the equation is –2,
(b) the sum of roots of the equation is 1 .
3
5. Given 3hx2 – 7kx + 3h = 0 has two real and equal roots, where h and k are positive. Find the
ratio h : k and solve the equation. PL3
6. Find the range of values of x for x2 – 7x + 10 . 0 and x2 – 7x < 0. Hence, solve the inequality
–10 , x2 – 7x < 0. PL5
7. The diagram on the right shows the graph of quadratic function
f(x) = – 1 [(x + p)2 + q]. The line y = 4 is the tangent to the
3
curve. Find PL3
(a) the roots for f(x) = 0,
(b) the values of p and of q,
(c) the equation of the axis of symmetry of the curve,
(d) the range of values of x when f(x) is positive.
8. The diagram on the right shows the graph of f(x) = x2 + bx + c,
where b and c are constants. Find PL3
(a) the values of b and of c,
(b) the coordinates of the minimum point,
(c) the range of values of x when f(x) is negative,
(d) the maximum value when the graph is reflected in the x-axis.
f(x)
y=4
0
3
x
7
f(x)
y = f(x)
0
2
6
9. A boat moves 24 km to the east and the water current was 3 km/h. The to-and-fro journey
took 6 hours. Find the velocity of the boat, in km/h, if the boat maintained its uniform
velocity. PL5
66
x
Quadratic Functions
10. An ancient Chinese book, Jiuzhang Suanshu which means “Nine Chapters on Mathematics
Arts” contained the following problem. PL4
CHAPTER 2
“ The height of a rectangular door is 6.8 units more than its width and the length
between two opposite vertices is 100 units. Find the width of the door.”
Using a quadratic formula, solve the problem.
11. The diagram on the right shows the cross section of a drain around
the house. If the shape of the drain is represented by the equation
1
y = x2 – 24x + 700, find PL5
5
(a) the width of the opening of the drain,
(b) the minimum depth of the drain.
12. The path of a shot put thrown by Krishna in a competition can be
represented by the quadratic function graph as shown in the diagram
on the right. The shot put is thrown at a height of 2 m and the path
passes through the maximum point (3, 2.5). PL4
2
(a) Express the equation of the path of shot put in the form
2
y = a(x – h) + k, where a, h and k are constants.
0
(b) Find the maximum distance of the horizontal throw, in
metres, by Krishna.
Exploring
House wall
Floor
Drain
y
(3, 2.5)
x
MATHEMATICS
The functions for three different water spouts in the shape of a
parabola in a pool are as follows.
Water spout I: h = –3d 2 + 4
Water spout II: h = –3.5d 2 + 3
Water spout III: h = –0.5d 2 + 1
For each of the functions, h metres represents the height of the water spout and d metres
is the horizontal distance of the water spout. Based on the given functions, answer the
following questions and explain your reason.
(a) Which water spout emits water from the highest point?
(b) Which water spout follow the narrowest path?
(c) Which water spout has the furthest distance?
67
Systems of
Equations
CHAPTER
CHAPTER 3
3
What will be learnt?
Systems of Linear Equations in Three
Variables
Simultaneous Equations involving One Linear
Equation and One Non-Linear Equation
List of
Learning
Standards
bit.ly/2pWDLl8
wORDS
KEY W
System of equations
Variable
Linear equation
Non-linear equation
Elimination method
Substitution method
Graphical method
68
Sistem persamaan
Pemboleh ubah
Persamaan linear
Persamaan tak linear
Kaedah penghapusan
Kaedah penggantian
Perwakilan graf
The Monsopiad Cultural Village in Sabah has
successfully sold 30 adult tickets to foreign
tourists for RM1 100. The ticket price for a trekking
package is RM30, whereas the ticket price for an
optional package is RM45 and the ticket price
for a standard package is RM55. The number of
trekking package tickets sold was twice the sum
of the optional and standard package tickets.
How do you determine the number of tickets sold
for each package?
3.1.1
Systems of Equations
Know?
The solutions to systems of linear equations
in three variables can be derived using the
Gaussian elimination method. This method
was invented circa 1810 by Friedrich
Gauss, a German mathematician. This is
an alternative method if you do not have a
graphical calculator or software.
For further information:
bit.ly/2OCmQi5
SIGNIFICANCE
OF THIS CH
APTER
Engineers use systems of equations
to solve problems involving voltages,
currents and resistances.
Biomedical, chemical, electrical,
mechanical and nuclear engineers
use systems of equations to derive the
measurements of solids and liquids.
Scan the QR code to watch
the video on the tradisional
dance of the Kadazandusun
tribe at the Monsopiad
Cultural Village.
3.1.1
bit.ly/2FNZjXk
69
CHAPTER 3
Did you
3.1
Systems of Linear Equations in Three Variables
Describing systems of linear equation in three variables
CHAPTER 3
Have a look at the prices of the packages offered by a
cinema in the advertisement on the right. How do you
determine the price of one ticket, one bottle of drink and
one bucket of popcorn?
EXCLUSIVE PACKAGE
SUTERA CINEMA
Three linear equations can be formed using the
variables x, y and z to represent the prices of one ticket,
one bottle of drink and one bucket of popcorn respectively.
x + y + 2z = 13
2x + 2y + z = 17
3x + 3y + 2z = 27
These linear equations formed are known as systems
of linear equations in three variables. A system of linear
equations refers to the presence of at least two linear
equations which contain the same set of variables. The
general form of a linear equation in three variables can be
written as follows:
ax + by + cz = d, where a, b and c are not equal to zero.
FLASHBACK
A linear equation is an
equation where the power
of the variable is 1.
Let’s see how systems of linear equations in three variables can be expressed in
three-dimensional planes.
INQUIRY 1
In groups
21st Century Learning
Aim: To describe systems of linear equations in three variables
Instructions:
1. Scan the QR code or visit the link on the right.
bit.ly/2ItuOpP
2. Click on all three boxes to display all three planes for the
equations x – 3y + z = 4, x – 3y + 3z = 4 and x – 3y + 3z = 0.
3. Observe the three planes.
4. Discuss with your groupmates your observations and record your findings in a sheet
of paper.
5. Each group shall move to other groups to compare the results obtained.
70
3.1.1
Systems of Equations
From the results of Inquiry 1, it is found that
A system of linear equations in three variables has three axes, namely the x-axis, y-axis
and z-axis. All three linear equations form a plane on each axis.
INQUIRY 2
In groups
21st Century Learning
Instructions:
bit.ly/35b4J8R
1. Scan the QR code or visit the link on the right.
2. Click on both boxes to display the two straight lines.
3. Observe both lines and record your group’s findings in a sheet of paper.
4. Compare your group’s findings with the results of Inquiry 1.
5. Present the comparison in front of the class.
From the results of Inquiry 2, it is found that there are only two axes, namely the x-axis and y-axis.
Every linear equation in two variables forms a straight line on each axis.
In general, a linear equation in two variables can be written in the form ax + by = c,
where a, b and c are constants. Geometrically, when a linear equation in two variables is
sketched on a plane, it will form a straight line as shown in Diagram 1(a).
Meanwhile, linear equations in three variables can be written in the form ax + by + cz = d,
where a, b, c and d are constants. When sketched, a three-dimensional plane will be formed,
as shown in Diagram 1(b).
y
z
ax + by = c
x
0
Diagram 1(a)
0
x
y
Diagram 1(b)
Example 1
Describe whether the following equations are systems of linear equations in three variables or not.
(a) 2x + 4y – z = 10
(b) p + 8q – 4r = 2
2
x + y = 10z 2(p + 6r) + 7q = 0
5y – z – 2x = 3 10r + p = 5q
Solution
(a) No, because there is an equation in which the highest power of the variable is 2.
(b) Yes, because all three equations have three variables, p, q and r, of power 1.
3.1.1
71
CHAPTER 3
Aim: To compare systems of linear equations in two variables and three
variables
Self Practice 3.1
1. Form linear equations in three variables for the following statement.
CHAPTER 3
Aiman bought 3 pairs of trousers, 2 shirts and a pair of shoes.
He spent RM750 for all his purchases.
2. Explain whether the following equations are systems of linear equations in three variables.
(c) 7a – c = 6b
(b) e(12 – 6g) = f 2
(a) 2m + 6(n – 2p) = 4
n = 5m + p
8e + 6 – 2f – 9g = 0 3 – 4c = 10a + b
2m
a + 3b = 2(c + b)
4m + p =
17f + e = 6 + 2e
5
6
Solving problems of linear equations in three variables
INQUIRY 3
In groups
Aim: To solve systems of linear equations in three variables
Instructions:
1. Scan the QR code or visit the link on the right.
bit.ly/31Z1gZe
2. Click on all three boxes to display all three planes for the linear
equations 2x + y + z = 3, –x + 2y + 2z = 1 and x – y – 3z = –6.
3. Do the planes intersect with each other? Determine the point of intersection of its
coordinates (x, y, z) between all three planes.
4. Determine whether the point of intersection (x, y, z) is the solution to all three linear
equations.
5. Record the opinion of each group member with regards to the relation between the
point of intersection and the solution to all the linear equations. Discuss accordingly.
From the results of Inquiry 3, the point of intersection between the three planes is the solution
to all three linear equations. In this case, there is only one solution because the planes only
intersect at one point.
INQUIRY 4
In groups
Aim: To solve systems of linear equations in three variables
Instructions:
1. Scan the QR code or visit the link on the right.
2. Click on all three boxes to display all three planes for the linear
equations x – 2y = 4, 2x – 3y + 2z = –2 and 4x – 7y + 2z = 6.
bit.ly/2Vk8OTC
3. Do the planes intersect with each other? Determine the point of intersection.
4. Record the opinion of each group member with regards to the relationship between the
point of intersection and the solution to all the linear equations. Discuss accordingly.
72
3.1.1
3.1.2
Systems of Equations
From the results of Inquiry 4, the three planes intersect along a straight line. Hence, this
system of linear equations has infinite solutions.
INQUIRY 5
In groups
From the results of Inquiry 5, it is found that the planes of all
three linear equations do not intersect at any point. Hence, this
system of linear equations has no solution.
The results of Inquiry 3, 4 and 5 show that there are three
types of solutions to systems of linear equations in three variables
as shown in the diagram below.
bit.ly/2IoqoRh
BRAINSTORMING
Using the GeoGebra
software, determine
the systems of linear
equations in three
variables which have
(a) only one solution
(b) infinite solutions
Solutions to systems of linear equations
in three variables
FLASHBACK
One
solution
Infinite
solutions
No
solution
The planes
intersect at
only one point
The planes
intersect along
a straight line
The planes
do not intersect
at any point
Systems of linear equations in three variables can be solved
by finding the values of all variables so that they satisfy all
three linear equations. The methods for solving systems of linear
equations in three variables are the elimination and substitution
methods.
The steps for solving systems of linear equations in three
variables through the elimination and substitution methods are
similar to the method for solving simultaneous equations in two
variables.
3.1.2
Simultaneous equations
in two variables can be
solved using the graphical
method, substitution
method or elimination
method.
FLASHBACK
There are three types of
solutions to simultaneous
equations in two variables.
When both lines are:
• Intersecting, there is a
unique solution.
• Parallel, there is no
solution.
• Overlapping, there are
infinite solutions.
73
CHAPTER 3
Aim: To solve systems of linear equations in three variables
Instructions:
1. Scan the QR code or visit the link on the right.
2. Click on all three boxes to display all three planes for the linear
equations 2x – 4y + z = 3, 4x + 8y + 2z = 14 and x – 2y + 0.5z = –1.
3. Are there any intersecting points among the three planes?
4. Record the opinion of each group member and discuss accordingly.
Quick Count
Example 22
Example
Solve the following system of linear equations using the
elimination method.
4x – 3y + z = –10
2x++yy++3z
3z==00
2x
–x++2y
2y––5z
5z==17
17
–x
CHAPTER 3
Solution
Choose any two equations.
4x – 3y + z = –10 … 1
… 2
2x + y + 3z = 0
Multiply equation 2 with 2 so that the coefficients of x are equal.
… 3
4x + 2y + 6z = 0
2 × 2:
… 3
4x
2y + 6z =x 0by subtracting
2 × 2:
Eliminate
the+ variable
1 from 3.
… 4
5y + 5z = 10
3 – 1:
… 4
+ 5z
10equations.
3 – 1:
Choose
another5yset
of =
two
… 5
2x + y + 3z = 0
… 5
2x
+
y
+
3z
=
0
…
–x + 2y – 5z = 17
6
… 6
–x + 2y – 5z = 17
Multiply equation 6 with 2 so that the coefficients of variable
x are equal.
… 7
6 × 2: –2x + 4y – 10z = 34
… 8
5y – 7z = 34
5 + 7:
12z = –24
4 – 8:
z = –2
Substitute z = –2 into 8 .
5y – 7(–2) = 34
5y + 14 = 34
5y = 20
y =4
Substitute y = 4 and z = –2 into 1 .
4x – 3(4) + (–2) = –10
4x – 12 – 2 = –10
4x = 4
x =1
Thus, x = 1, y = 4 and z = –2 are the solutions to this system of
linear equations.
Determine the solution
to Example 2 using a
scientific calculator.
1. Press MENU
2. Press  twice
3. Press ALPHA (–)
4. Press 1 for Simul
Equation
5. Press 3
6. Key in the coefficients of
x, y and z.
Press 4 = , –3 = ,
1 = , –10 = , 2 = ,
1 =, 3 =, 0 =,
–1 = , 2 = , –5 = ,
17 =
7. The screen will show:
x
1
Press =
y
4
Press =
z
–2
Thus, x = 1, y = 4 and z = –2.
Example 33
Example
Solve the following system of linear equations using the
substitution method.
3x – y – z = –120
2z ==30
30
yy––2z
180
xx++yy++zz ==180
74
3.1.2
Systems of Equations
… 1
3x – y – z = –120
… 2
y – 2z = 30
… 3
x + y + z = 180
Express z in terms
From 1 , z = 3x – y + 120 … 4
of x and y
Substitute 4 into 2 .
y – 2(3x – y + 120) = 30
y – 6x + 2y – 240 = 30
– 6x + 3y = 270
Express y in terms
y = 90 + 2x … 5
of x
Substitute 4 and 5 into 3 .
x + (90 + 2x) + [3x – (90 + 2x) + 120] = 180
x + 2x + 3x – 2x + 90 – 90 + 120 = 180
4x = 60
x = 15
Substitute x = 15 into 5 .
y = 90 + 2(15)
= 120
Substitute x = 15 and y = 120 into 3 .
15 + 120 + z = 180
z = 45
Thus, x = 15, y = 120 and z = 45 are the solutions to this
system of linear equations.
QR
The Gaussian elimination
method can also be
used to solve systems
of linear equations. Visit
the following link for
more information on the
Gaussian elimination
method.
CHAPTER 3
Solution
bit.ly/2nA2g6U
Mind Challenge
Solve Example 3 using the
elimination method. Did
you get the same solution?
Example 4
Solve the following system of linear equations.
x – y + 3z = 3
–2x + 2y – 6z = 6
y – 5z = –3
Solution
x – y + 3z = 3
–2x + 2y – 6z = 6
y – 5z = –3
…
…
…
1
2
3
Since equation 3 only has two variables, which are y and z,
then the variable x in equations 1 and 2 need to be eliminated.
1 × 2: 2x – 2y + 6z = 6 … 4
4 + 2 : 0 + 0 + 0 = 12
0 = 12
QR
Solution to Example 4
using the GeoGebra
software.
bit.ly/2OwUo0W
Thus, this system of linear equations has no solution as 0 ≠ 12.
3.1.2
75
Example 5
Solve the following system of linear equations.
3x + 5y – 2z = 13
–5x – 2y – 4z = 20
–14x – 17y + 2z = –19
CHAPTER 3
Solution
… 1
3x + 5y – 2z = 13
… 2
–5x – 2y – 4z = 20
–14x – 17y + 2z = –19 … 3
Multiply equation 1
… 4
with 2 to eliminate the
1 × 2: 6x + 10y – 4z = 26
variable z
… 5
11x + 12y = 6
4 – 2:
… 6
1 + 3 : –11x – 12y = – 6
+
:
0
+
0
=
0
5
6
0=0
Thus, this system of linear equations has infinite solutions because 0 = 0.
Self Practice 3.2
1. Solve the following systems of linear equations using the elimination method.
(a) 7x + 5y – 3z = 16
(b) 4x – 2y + 3z = 1
3x – 5y + 2z = –8 x + 3y – 4z = –7
5x + 3y – 7z = 0 3x + y + 2z = 5
2. Solve the following systems of linear equations using the substitution method.
(a) 2x + y + 3z = –2
(b) 2x + 3y + 2z = 16
x – y – z = –3 x + 4y – 2z = 12
3x – 2y + 3z = –12 x + y + 4z = 20
Solving problems involving systems of linear equations in three
variables
Example 6
MATHEMATICS APPLICATION
Tommy has three condominium units, type A with 1 bedroom,
type B with 2 bedrooms and type C with 3 bedrooms. All
three condominium units are rented and the total rent received
by Tommy is RM1 240 per day. Tommy needs to save 10% of
the rent of the type A unit, 20% of the rent of the type B unit,
and 30% of the rent of the type C unit to pay for maintenance
charges. The total daily savings is RM276. The rent for the
type C unit is twice the rent for the type A unit. What is the
daily rent for each condominium unit owned by Tommy?
76
3.1.2
3.1.3
Systems of Equations
Solution
u
u
u
u
Total rent is RM1 240 per day.
Savings for maintenance charges:
• Type A unit is 10% of the rent.
• Type B unit is 20% of the rent.
• Type C unit is 30% of the rent.
Total daily savings in RM276.
The rent for the type C unit is twice
the rent for the type A unit.
4 . Making a conclusion
Total rent per day
= 280 + 400 + 560
= RM 1 240
Total daily savings
= 0.1(280) + 0.2(400) + 0.3(560)
= 28 + 80 + 168
= RM276
2 . Planning a strategy
Form three equations which involve
the total daily rent, total daily savings
and the rent of the type C unit.
u Let the rent of the type A unit be a,
rent of the type B unit be b and rent of
the type C unit be c.
u
3 . Implementing the strategy
a + b + c = 1 240 … 1
… 2
0.1a + 0.2b + 0.3c = 276
… 3
c = 2a
2 × 10: a + 2b + 3c = 2 760 … 4
1 × 2: 2a + 2b + 2c = 2 480 … 5
… 6
– a + c = 280
4 – 5 :
Substitute 3 into 6 .
– a + 2a = 280
a = 280
Substitute a = 280 into 3 .
c = 2(280)
= 560
Substitute a = 280 and c = 560 into 1 .
280 + b + 560 = 1 240
840 + b = 1 240
b = 400
The rent for the type A condominium
unit is RM280, type B condominium unit
RM400 and type C condominium unit
RM560.
Self Practice 3.3
1. Patricia invested RM24 500 in three unit trusts. She divided the money into three different
unit trust accounts, P, Q and R. At the end of the year, she obtained a profit of RM1 300. The
annual dividends for the accounts are 4%, 5.5% and 6% respectively. The final amount of
money in account P is four times that in account Q. How much money did she invest in each
unit trust account?
2. Billy Restaurant ordered 200 stalks of flowers in conjunction with Mother’s Day. They
ordered carnations which cost RM1.50 per stalk, roses which cost RM5.75 per stalk and
daisies which cost RM2.60 per stalk. Of the three types of flowers ordered, the number of
carnations was the largest. Meanwhile, the number of roses was 20 stalks less than that of
daisies. The total price of the flowers is RM589.50. How many flowers of each type were
ordered?
3.1.3
77
CHAPTER 3
1 . Understanding the problem
3. Ramasamy wants to buy pens, pencils and notebooks for the new school term. He has RM102
to spend. The price of a pen is RM5, a pencil RM3, and a notebook RM9. Ramasamy intends
to spend the same amount of money on pens and pencils. The total number of pens and
pencils to be purchased must be equal to the number of notebooks to be purchased. How
many of each item will he purchase? Write a system of equations to solve this problem.
CHAPTER 3
Intensive Practice 3.1
Scan the QR code or visit bit.ly/2IEEJZT for the quiz
1. Form systems of linear equations for the following situations and answer the questions given.
(a) The sum of angles in a triangle is 180°. The largest angle is 20° more than the sum of the
other two angles and is 10° more than three times the smallest angle. What is the measure
of each angle of the triangle?
(b) The sum of three numbers is 19. If the first number is multiplied by 2, the sum of the
three numbers is 22, and if the second number is multiplied by 2, the sum becomes 25.
Find the value of each number.
2. Solve the following equations using the elimination and substitution methods.
(a) x + y + z = 3
(b) 2x + y – z = 7
(c) x + y + z = 3
x + z = 2 x – y + z = 2 2x + y – z = 6
2x + y + z = 5 x + y – 3z = 2 x + 2y + 3z = 2
(d) 2x – y + z = 6
(e) x + y + 2z = 4
(f) x + 2y + z = 4
3x + y – z = 2 x + y + 3z = 5 x – y + z = 1
x + 2y – 4z = 8 2x + y + z = 2 2x + y + 2z = 2
3. A bakery bakes three types of bread, with the monthly cost being RM6 850 for 2 150 loaves
of bread. The cost of baking a loaf of butterscotch bread is RM2, a chocolate bread RM3 and
a coconut bread RM4. The sale prices of a loaf of butterscotch bread, a chocolate bread and
a coconut bread are RM3, RM4.50 and RM5.50 respectively. If the bakery makes a profit of
RM2 975 monthly, how many loaves of each type of bread will it bake?
4. Andrea sells flower pots of different sizes. The price of a small flower pot is RM10, medium
flower pot RM15 and large flower pot RM40. Every month, the number of small flower pots
sold is equal to the total number of the medium and large flower pots sold. The number of
medium flower pots sold is twice the number of large flower pots sold. Andrea needs to pay a
rent of RM300 per month for her business premise. What are the minimum numbers of pots
of each size which Andrea has to sell in a month so that she can pay the monthly rent?
5. Mr Chong intends to purchase a few chickens, rabbits and ducks for his farm. The total
number of animals is 50. He has RM1 500 to be spent. A chicken costs RM20, a rabbit RM50
and a duck RM30. The number of chickens is equal to the number of ducks. How many of
each animal should Mr Chong purchase? Write a system of linear equations to solve this
problem.
78
3.1.3
Systems of Equations
3.2
Simultaneous Equations Involving One Linear Equation and
One Non-Linear Equation
Solving simultaneous equations involving one linear equation and
one non-linear equation
INQUIRY 6
In groups
21st Century Learning
CHAPTER 3
Aim: To identify simultaneous equations
Instructions:
1. Form a few groups of three to four members each.
2. Read the following statements and construct the appropriate equations.
STATEMENT 1
Chong has a rectangular garden. The length of fencing
which will be used to fence the garden is 200 m. The
area of the garden is 2 400 m2. What is the length and
width of the garden?
STATEMENT 2
Shida is sewing a rectangular tablecloth. The perimeter
of the tablecloth is 800 cm and the area is 30 000 cm2.
Find the length and width of the table.
STATEMENT 3
The difference between two numbers is 9 and the product of the two numbers is
96. Find the values of the numbers.
3. Answer the following questions:
(a) How many equations can be formed in each statement?
(b) How many variables are involved?
4. Discuss with your group members and record your findings on a piece of paper.
5. Each group shall elect a representative to present their findings to the class.
6. Other group members can ask questions to the elected representatives.
7. Repeat steps 5 and 6 until all groups have finished presenting.
From the results of Inquiry 6, each of the three statements had two equations of two
variables, namely a linear equation and non-linear equation. What are the characteristics
which distinguish linear equations from non-linear equations? How do we solve simultaneous
equations which involve a linear equation and non-linear equation?
3.2.1
79
InQuirY 7
Individual
21st Century Learning
CHAPTER 3
Aim: To explore the point of intersection between linear and non-linear equations
Instructions:
1. Scan the QR code or visit the link on the right.
2. Click on both boxes to display the shapes of graphs for the equations
x + 2y = 10 and y2 + 4x = 50.
bit.ly/2OqKBK7
3. What can you conclude about the shapes of the two graphs?
From the results of Inquiry 7, the point of intersection between the graph of linear equation
x + 2y = 10 and the non-linear equation y2 + 4x = 50 is the solution to both equations. The
solution to both equations is also known as the solution of simultaneous equations.
Solving simultaneous equations means finding the values of the variables which
satisfy the equations concerned. These simultaneous equations can be solved using
the elimination method, substitution method or graphical representation method.
Example 7
Solve the following simultaneous equations using the
substitution method.
2x + y = 4
y2 + 5 = 4x
Solution
From
1
,
2x + y = 4
y2 + 5 = 4x
Substitute
2x = 4 – y
4–y
x=
2
1
…
3
2
Solve Example 7 when y is
expressed in terms of x for
the linear equation 2x + y = 4.
Did you get the same
solutions?
2
Make x the subject of the
formula
.
4–y
y +5 =4
2
y2 + 5 = 8 – 2y
y2 + 2y – 3 = 0
Solve the quadratic equation by
(y + 3)(y – 1) = 0
factorisation
y = –3 or y = 1
Substitute y = –3 and y = 1 into 3 .
4 – (–3)
4–1
x=
or
x=
2
2
7
3
=
=
2
2
3
2
into
…
…
Mind Challenge
( )
FLASHBACK
Quadratic equations
can be solved using the
following methods:
(a) Factorisation
(b) Formula
(c) Completing the square
7
3
, y = –3 and x = , y = 1 are the solutions to these
2
2
simultaneous equations.
Thus, x =
80
3.2.1
Systems of Equations
Example 8
Solve the following simultaneous equations using the elimination method.
2x + y = 4
x2 – 2xy = 3
Solution
2x + y = 4 … 1
x – 2xy = 3 … 2
1 × 2x: 4x2 + 2xy = 8x … 3
5x2 = 3 + 8x
2 + 3:
2
5x – 8x – 3 = 0
–b ± !w
b2 – 4ac
Use the quadratic
x=
formula
2a
2
8 ± !w
(–8) – 4(5)(–3)
=
2(5)
x = 1.9136 or x = – 0.3136
Substitute x = 1.9136 into 1 .
2(1.9136) + y = 4
3.8272 + y = 4
y = 0.1728
Substitute x = – 0.3136 into 1 .
2(– 0.3136) + y = 4
– 0.6272 + y = 4
y = 4.6272
Thus, x = 1.9136, y = 0.1728 and x = – 0.3136, y = 4.6272 are the solutions to these
simultaneous equations.
CHAPTER 3
2
Example 9
Solve the following simultaneous equations using the graphical representation method.
2x + y = 4
y2 – 4x = 0
Solution
Construct a table to determine the points which need to be plotted.
x
0
1
2
3
4
5
Value of y in the equation
2x + y = 4
4
2
0
–2
–4
–6
Value of y in the equation
y2 – 4x = 0
0
±2
±2.8
±3.5
±4
±4.5
3.2.1
81
Construct a graph based on the values in the table.
y
Solve the following
simultaneous equations
2x + y = x2 – xy + y2 = 7.
y2 – 4x = 0
4
2
CHAPTER 3
0
(1, 2)
2
4
6
8
x
Tech Whizz
–2
–4
–6
Mind Challenge
The solution to Example 9
using the Desmos software
(4, – 4)
2x + y = 4
Based on the graph above, there are two points of intersection
which represent the solutions to both equations. Thus, the
solutions to these simultaneous equations are (1, 2) and (4, –4).
bit.ly/2KHpcbw
Self Practice 3.4
1. Solve the following simultaneous equations using the elimination, substitution or graphical
representation method.
(a) 2x – y = 7
(b) 5y + x = 1
(c) y = 3 – x
1 1
y2 – x(x + y) = 11 x + 3y2 = –1
– =2
x y
(e) 2x + 4y = 9
(f) x + y – 4 = 0
(d) 3x + 5y = 1
4
x + 2y = 4x2 + 16y2 = 20x + 4y – 19 x2 – y2 – 2xy = 2
y
2. Solve the following simultaneous equations using the graphical representation method.
(a) Draw the graph for the following pairs of equations for the domain –5 < x < 5.
Hence, determine the solution to the simultaneous equations.
2y – x = 1
xy + x2 = 26
(b) Draw the graph for the following pairs of equations for the domain –3 < x < 4. Hence,
determine the solution to the simultaneous equations.
x–y=2
4x2 + 3y2 = 36
82
3.2.1
Systems of Equations
Solving problems involving simultaneous equations
MATHEMATICS APPLICATION
A food packaging factory intends to pack dodol in a container in the
shape of a right prism with a square base, as shown in the diagram.
Given that the total length of sides of the right prism is 133 cm and
ED = BC = 25 cm. Can a piece of dodol of volume 600 cm3 be packed into
the container? Justify your answer.
1 . Understanding the problem
Container in the shape of a right prism
with a square base.
u Total length of sides of the container
= 133 cm
u ED = BC = 25 cm
u To determine whether dodol of volume
600 cm3 can be packed into the
container.
u
F
Let the length of side of the base be x
and the height be y.
u Form a non-linear equation for the
length of BC.
u Form a linear equation for the total
length of sides of the prism.
u Volume of prism
= area of cross-section × height
u
4 . Making a conclusion
Volume of container = 588
1
× 7 × 24 × x = 588
2
x = 7 cm
2
D
C
3 . Implementing the strategy
x2 + y2 = 252
5x + 2y + 50 = 133
5x + 2y = 83
From 2 ,
83 – 5x
y=
2
Substitute 3 into 1 .
…
1
…
2
…
3
(83 2– 5x) = 25
6 889 – 830x + 25x
x +(
) = 625
4
2
x2 +
2
2
2
2 . Planning a strategy
3.2.2
B
A
Solution
Substitute x = 7 into equation
5(7) + 2y = 83
y = 24 cm
E
CHAPTER 3
Example 10
4x2 + 25x2 – 830x + 6 889 – 2 500 = 0
29x2 – 830x + 4 389 = 0
(29x – 627)(x – 7) = 0
627
or x = 7
x=
29
627
into 3 .
Substitute x =
29
627
83 – 5
29
y=
2
364
= –
(Ignored)
29
Substitute x = 7 into 3 .
83 – 5(7)
y=
2
= 24
1
Volume of container = × 7 × 7 × 24
2
= 588 cm3
Hence, a piece of dodol of volume 600 cm3
cannot be packed into the container
because the volume of container is only
588 cm3.
( )
83
Self Practice 3.5
1. Audy cuts out a rectangular plank of area 72 cm2 and perimeter 34 cm. Calculate the length
and width of the plank.
xm
ym
Pond
CHAPTER 3
2. The diagram shows the plan of a rectangular garden
which will be constructed by Syarikat Pesona Alam. The
corner of a garden will have a rectangular pond. The area
to be covered with grass is 96 m2 and the perimeter of the
pond is 20 m. Calculate the value of x and of y.
Grassy area
10 m
12 m
Intensive Practice 3.2
Scan the QR code or visit bit.ly/2IDhbEE for the quiz
1. Solve the following simultaneous equations.
(b) k – 3p = –1
(a) x – 3y + 4 = 0
x2 + xy – 40 = 0 p + pk – 2k = 0
2y
2. Find the coordinates of the points of intersection of the curve x – = 1 and straight line
y x
2x + y = 3.
3. Given that (–2, 2) is the solution to the following simultaneous equations:
x + 1 y = h and 1 + 2 = k
2
2
x y
Find the value of h and of k. Hence, find another solution.
4. The hypothenuse of a right-angled triangle is (2x + 3) cm. The lengths of two other sides are
x cm and (x + y) cm respectively. Given that the perimeter of the triangle is 30 cm, find the
value of x and of y.
5. Given that the total surface area of a cuboid with a square base is 66 cm2 and the total length
of the sides is 40 cm. Find the possible volumes of the cuboid.
6. A fish is moving in a circular manner, with the equation of its locus being
2x2 + 11y2 + 2x + 2y = 0. A boat is moving along a straight line of equation x – 3y + 1 = 0
and it intersects with the circular locus of the fish. Find the points of intersection between the
locus of the fish and trajectory of the boat.
7. A sailing boat is moving in a circular manner, with the
equation of its locus being 2x2 + 4y2 + 3x – 5y = 25.
Meanwhile, a speedboat is moving along a straight line of
equation y – x + 1 = 0 and it intersects with the locus of
the sailing boat. Find the points of intersection between the
trajectories of the sailing boat and the speedboat.
84
3.2.2
Systems of Equations
Summary OF Chapter 3
Systems of Linear
Equations in Three
Variables
SYSTEMS OF
EQUATIONS
a � � l�
No solution
Solving problems
in daily life
Simultaneous Equations
• one linear equation
• one non-linear equation
Elimination
method
u��n�
Substitution
method
t��e��o�
u��o��
�s��
�� �
Satu
Infinite
penyelesaian
solutions
�a��o ��
u��n�
Elimination
method
i��o��
e�
CHAPTER 3
i��o��e�
Substitution
method
One solution
Graphical
method
WRITE YOUR JOURNAL
Think of a problem in your surroundings which can be solved using systems of linear and
non-linear equations. Formulate the problem in the form of a system of linear equations with
proper definitions for the variables used. State the relation between the variables. Solve the
system of equations which has been constructed. Then, write a report on the problem and
display it to the class.
85
CHAPTER 3
MASTERY PRACTICE
1. Form systems of linear equations in three variables for the following situations. PL1
(a) Abdullah buys a History book, two Mathematics books and three Science books for
RM120. Chong buys two History books, three Mathematics books and two Science
books for RM110. Meanwhile, Kaladevie buys a History book, four Mathematics books
and two Science books for RM180.
(b) There are a total of 30 coins which consist of 10 cents, 20 cents and 50 cents coins in a
box. The total value of the coins is RM20.60. Salmah bought an ice cream using two 50
cents and three 20 cents coins.
2. Solve the following systems of linear equations. PL2
(a) x – y + 2z = 3
(b) x + 2y + 5z = –17
x + y – 3z = –10 2x – 3y + 2z = –16
2x + y – z = –6 3x + y – z = 3
3. The second angle of a triangle is 50° less than four times the first angle. The third angle is
40° less than the first angle. Find the value of each angle in the triangle. PL3
4. Given that (5, h) is one of the solutions to the following simultaneous equations.
h(x – y) = x + y – 1 = hx2 – 11y2
Find the value of h and the other solution to the simultaneous equations.
PL4
5. Every month, Raju receives sources of income from his fixed salary as a sales officer, house
rental and online sales. His total monthly salary is RM20 000. If RM500 is added to his
monthly salary, it will be twice the total income from house rental and online sales. The
total monthly salary and online sales income is twice the house rental income. How much
does Raju receive from each source of income every month? PL4
6. Encik Abu plants vegetables on a plot of land in the shape of a right-angled triangle. Given
that the longest side of the land is p metres, the other two sides are q metres and (2q – 1)
metres respectively. Encik Abu fenced the land using a fencing of length 40 metres. Find
the length, in metres, of each side of the land. PL4
7. Prove that a straight line passing through (0, –3) intersects a curve x2 + y2 – 27x + 41 = 0
at point (2, 3). Does the straight line intersect the curve at any other point? Justify your
answer. PL4
8. A piece of wood measures y cm in length and 3x cm in width. A worker intends to cut the
piece of wood into two small triangular pieces of wood. The perimeter of each triangle is
24 cm and the longest side of either triangle is (x + y) cm. Calculate the area, in cm2, of the
original piece of wood. PL4
86
Systems of Equations
9. The diagram on the right shows the plan of a
rectangular room. A rectangular carpet is to be placed
with a distance of 1 m from the wall of the room. The
area and perimeter of the carpet are 8.75 m2 and 12 m
respectively. Find the length and width, in m, of the
room. PL4
1m
1m
1m
1m
P
T
y cm
S
11. Mr. Chee Hong instructed the students of Form 4
Kembara to draw a rectangular mural of length 7x m
and width y m on the wall of the canteen. Two different
shapes will be drawn on the wall as shown in the diagram
on the right. AED is a semicircle. Given that the area of
the wall is 28 m2 and the perimeter of ABCDE is 26 m,
find the diameter and radius of the semicircle. PL5
Exploring
Q
28x cm
R
D
A
ym
E
B
7x m
C
MATHEMATICS
Encik Awang, a chemist, has three types of solutions. Every day, he will prepare a few
solutions of specific proportions. One day, Encik Awang intends to prepare a solution from
three types of solutions. The first solution must contain 10% of acid, the second solution
40% of acid and the third solution 60% of acid. Encik Awang intends to prepare 1 000 litres
of mixed solution with an acid concentration of 45%. The available amount of 40% acid
solution is twice the amount of 10% acid solution. How much of each solution would you
suggest Encik Awang use?
1.
Write three equations based on the above statements.
2.
Write your suggested workings to Encik Awang.
87
CHAPTER 3
10. The diagram on the right shows a rectangular piece of
cardboard PQRS of area 224 cm2. A semicircle STR was
cut out from the cardboard. Given that the perimeter of
the remainder of the cardboard is 72 cm, find the value
of x and of y. PL4
CHAPTER
CHAPTER 4
4
Indices, Surds
and Logarithms
What will be learnt?
Laws of Indices
Laws of Surds
Laws of Logarithms
Applications of Indices, Surds and Logarithms
List of
Learning
Standards
bit.ly/2VyB2Kk
wORDS
KEY W
Index
Base
Rational number
Irrational number
Surd
Radical
Recurring decimal
Conjugate surd
Logarithm
Natural logarithm
Algebraic expression
Coefficient
88
Indeks
Asas
Nombor nisbah
Nombor tak nisbah
Surd
Radikal
Perpuluhan berulang
Surd konjugat
Logaritma
Logaritma jati
Ungkapan algebra
Pekali
4.1.1
Indices, Surds and Logarithms
Did you Know?
For further information:
CHAPTER 4
The population of the country needs to
be estimated in order to plan the future of
the nation. Knowledge on the growth rate
of the Malaysian population will enable
preparations to be made in various aspects,
including medical facilities for children, new
registrations for year 1 students and so on.
In your opinion, how can the population of
Malaysia in a particular year be estimated?
John Napier was a famous Scottish
mathematician who introduced logarithms.
Logarithms is a mathematical tool
which simplifies calculations, especially
multiplications which are frequently utilised
in astronomy.
bit.ly/2pgYYWz
SIGNIFICANCE
OF THIS CH
APTER
The half-life of a radioactive substance is
given by the function N(t) = N0e–λt, where N0 is
the initial mass of the radioactive substance,
N(t) is the remaining mass of the radioactive
substance post-decay, t is the time of decay
and λ is the decay constant. By substituting
the values of N0, N(t) and λ into the function,
physicists can determine the time of decay of
a radioactive substance.
Biologists can determine the growth rate of
bacteria from time to time if the bacteria were
allowed to proliferate.
The intensity of an earthquake can be
determined by using exponential functions.
This enables geoscientists to calculate its
magnitude using the Richter scale.
Scan this QR code to
watch a video on the
population of Malaysia.
bit.ly/2qWcv70
4.1.1
89
4.1
Laws of Indices
InQuirY 1
In pairs
21st Century Learning
Aim: To recall the laws of indices
Instructions:
1. Scan the QR code or visit the link on the right.
bit.ly/2onDQy8
2. Take note of the list of laws of indices. Cut out all shapes and paste
at the corresponding laws of indices in the table.
3. Write an example of index law using algebraic expressions as shown in the following
table.
CHAPTER 4
Index law
am × an
=
Example
am + n
t 2 × t 3 = t2 + 3 = t5
4. Display your partner’s and your findings.
5. Along with your partner, move around and observe other teams’ results.
Simplifying algebraic expressions involving indices
You have learnt that an is an index number with a being the base and n being the index.
How can an algebraic expression involving indices be simplified by using the laws of indices?
Let’s explore.
InQuirY 2
Individual
Aim: To simplify algebraic expressions involving indices
Instructions:
1. List the laws of indices which you have learnt.
2. Scan the QR code or visit the link on the right.
bit.ly/2MmURQD
3. Using the laws of indices you have listed down, simplify each of
the provided algebraic expressions.
4. Click on the “Check Answer” button to check your answers.
5. Discuss with your friends the ways by which you have obtained your answers.
From the results of Inquiry 2, we can deduce that:
An algebraic expression involving index numbers can be simplified by using the
laws of indices.
90
4.1.1
Indices, Surds and Logarithms
Example 1
Simplify the following algebraic expressions.
2n
m
(a) 4 ×n 4
4
(c) (5x −1)3 × 4xy2 ÷ (xy)− 4
m+2
m
(b) 3 m– 3
3
(d) 4a3b2 × (4ab3)– 4
Solution
2n
m
m+2
m
m
2
m
(a) 4 ×n 4 = 42n + m – n
(b) 3 m– 3 = 3 × 3m – 3
4
3
3
3m(32 – 1)
= 4n + m
=
3m
CHAPTER 4
=8
(d) 4a3b2 × (4ab3)– 4
(c) (5x –1)3 × 4xy2 ÷ (xy)– 4
(5x –1)3 × 4xy2
=
= 4a3b2 × 1 3 4
(xy)– 4
(4ab )
= 53x –3 × 4xy2 × (xy)4
4a3b2
=
256a4b12
= 125 × 4 × x –3 + 1 + 4 × y2 + 4
= 500x2y6
= 1 10
64ab
Example 2
Simplify the following algebraic expressions.
(a) a
–1
3
–1
× 2a
2
a2 × 2!w
a –3
(c) 3!w
Solution
(a) a
–1
3
–1
× 2a
2
–2
(b) 2a 3
–
a 2
3
1
–1
–1
(d) a 2 (a 2 + 2a 2 – 3a 2 )
–2
–3
(b) 2a 3 = 2a–2 ÷ a 2
–
a 2
– 1 + (– 1 )
–2 – – 3
= 2a ( 2 )
= 2a 3 2 –1
=2×a
3
–1
×a
2
–5
= 2a 6 = 2a
= 25 = 21
a6
a2
2
–3
a2 × 2!w
a –3 = a 3 × a
(c) 3!w
2
–1
2 + –3
= a 3 ( 2 ) =a
= a3
=a
4.1.1
–5
6
2
3
1
–1
(d) a 2 (a 2 + 2a 2 – 3a 2 )
2 – 3
2
–1
–1
2
3
–1+3
2
–1
1
–1
–1
× a 2 + a 2 × 2a 2 – a 2 × 3a
2
–1+1
+ 2a
2
= a = a1 + 2a0 – 3a–1
= 15 = a + 2 – 3
a
a6
2
2
–1–1
– 3a
2
2
91
Example 3
Show that
(a) 72x – 1 = 49
7
(b) 3x + 4 + 3x + 5 + 3x is divisible by 25 for all positive integers x.
x
Solution
(a) 72x – 1 = 7
7
x
49
=
7
(b) 3x + 4 + 3x + 5 + 3x = 3x(34) + 3x(35) + 3x
= 3x(81 + 243 + 1)
= 3x(325)
Since 325 is a multiple of 25, thus 3x + 4 + 3x + 5 + 3x is divisible by 25 for all positive
integers x.
CHAPTER 4
2x
Self Practice 4.1
1. Simplify the following algebraic expressions.
3x
x
b–2
b
(a) 5 ×–x 5
(b) 7 b +–3 7
5
7
(e) (xy2)3 × x3y5
(d) c4d3 × c3d5
(h) ( p2q−1)5 × q8
(g) (3x2y)3 × (x3)4 ÷ x16y2
(j) (49−2xy)3 ÷ (7xy)−2
(k) 20x−7y2 ÷ 4x3y−4
2. Simplify the following algebraic expressions.
1
3
–1
(a) a 3 × 2a 2
(b) 4a3
–
a 5
(c) 5!w
a7 × 4!w
a–9
3. Show that
(d) a
(a) 43a − 2 = 64
16
a
–3
2
(a
1
2
–3
+ 3a
2
a–3
a+4
(c) 9 + 9
81
(f) (7x−1)2 × (49−2xy)3
(i) ( pq5)4 × p3
(l) 6a7b−2 ÷ 36a3b−4
–5
– 3a
(b) 92a + 2 = 81(81a)
2
)
a
(c) 73a − 4 = 343
2 401
4. Show that 4x + 2 + 4x + 1 + 4x is divisible 7 for all positive integers x.
Solving problems involving indices
Equations involving indices can be solved as follows:
If am = an, then m = n or if am = bm, then a = b when
a > 0 and a ≠ 1.
92
4.1.1
4.1.2
Indices, Surds and Logarithms
MATHEMATICS
POCKET
Example 4
Solve each of the following equations.
(a) 32x = x1– 1
8
(b) a5 = 243
(c) 27(813x) = 1
• If 5x = 54, then x = 4.
• If x5 = 55, then x = 5.
Solution
(b) a5 = 243
= 35
a= 3
Compare the indices
Express in index form
Compare the indices
Mathematics
30 = 1
am × an = am + n
2. Press 9
 3 x£ 2

ALPHA ) .
3. Press SHIFT CALC .
4. Press = to obtain the
value of x.
Application
Husna has RM1 000 000. She invests that sum of
money into an investment institution that offers an
annual return of 6%. Husna's investment amount after
n years is calculated with the equation J = p(1 + k)n
using p as the investment at the beginning of the year
and k as the annual return rate. Determine Husna's
investment amount after 20 years.
4.1.2
Given 3x = 92x, find the value
3
of x.
1. Press 3 x£ ALPHA )
„ ALPHA CALC .
(c) 27(813x) = 1
33(34)3x = 30
33 + 12x = 30
3 + 12x = 0
12x = –3
x = – 3
12
= – 1
4
Example 5
Quick Count
Express both sides of equations in a same base
SAMPLE
93
CHAPTER 4
(a) 32x = x1– 1
8
25x = 2–3(x – 1)
5x = –3x + 3
8x = 3
x= 3
8
Solution
1 . Understanding the problem
◆
◆
◆
◆
◆
Initial investment, p is RM1 000 000
Return rate, k is 6% per annum
Investment formula, J = p(1 + k)n
n = 20
Determine the total investment after
20 years
2 . Planning a strategy
Substitute the values of k, p and n into
the investment formula.
3 . Implementing the strategy
CHAPTER 4
4 . Making a conclusion
When J = 3 207 135 and k = 0.06, then
3 207 135 = 1 000 000(1 + 0.06)n
3.207135 = (1.06)n
n = 20
Thus, n = 20 years.
J = p(1 + k)n
(
)
6 20
100
= 1 000 000(1 + 0.06)20
= 1 000 000(3.207135)
= 3 207 135
Thus, Husna's total investment is
RM3 207 135.
= 1 000 000 1 +
Self Practice 4.2
1. Solve the following equations:
(a) 4x – 1 = 8x + 3
(b) 3x + 3 – 3x + 2 = 2
(c) 8x – 3 = 4
64
2x
2. A ball was released at a height of h cm from the surface of the Earth. The ball will bounce
with 90% of its initial height after the ball hits the surface of the Earth. The height of the ball
after l bounces is given by h = 10 × (0.9)l. Determine the height of the ball, in cm,
(a) when the ball was released,
(b) after 10 bounces.
Intensive Practice 4.1
Scan the QR code or visit bit.ly/2IAEoaM for the quiz
1. Simplify each of the following:
y3(3zx)2
z4yx2
(b)
(a)
3
9x
zxy2
2 3
−2 2
(d) (ef ) ÷ (e f )
(e) 4.2x4y14 ÷ 0.6x9y5
(c) [(xy)5 × 2xy3]2
(f) (7x−1)2 × (49−2xy)3 ÷ (7xy)−2
2. If 2x – 2 = 2(16), find the value of x.
3. Solve 25x – 53x – 4 = 0.
4. Solve 4(2m + 1) – 16m = 0.
94
4.1.2
Indices, Surds and Logarithms
5. Find the way to the FINISH box by choosing the correct answers.
START
Solve
3
n–2
1
× 27 =
81
2
3x – 5
n
=
–2
3
Solve
5n + 1 – 5n + 5n – 1
= 105
Solve
Solve
1
25x + 2 =
625x
4 x + 3 – 4x + 2 = 3
1
2
Solve
4
2 ÷ 4x – 1 = 162x
4x + 1
1
2
2x – 1
Solve
1
2
= 64
16(3n – 1) = 27n
2
324x = 48x + 6
–1
Solve
Solve
Solve
x
CHAPTER 4
Solve
x+4
– 2x + 3 = 1
–3
Solve
162x – 3 = 84x
FINISH
6. In a research, a type of bacteria will multiply itself in 1 minute. The amount of bacteria at the
start of the research was 300. The amount of bacteria after t minutes is given by 300(3t).
(a) Determine the amount of bacteria after 9 minutes.
(b) Determine the time, t, in minutes for the amount of the bacteria to be 72 900.
k t
7. The population of country M can be estimated with the growth model, P = A 1 +
100
where P is the expected population, A is the population in year 2017, k is the growth rate and
t is the number of years after 2017. The population of this country in 2017 was approximately
30 million. Assuming that this population increases with a rate of 3% each year, estimate the
population of this country in the year 2050.
(
)
8. Mr. Prakesh invested RM20 000 in a bank with an annual interest rate of 10%. Mr. Prakesh's
investment amount after t years can be determined by the formula P = f(1 + r)t where f is the
initial investment value and r is the annual return rate. Determine Mr. Prakesh's investment
amount after 10 years.
95
4.2
Laws of Surds
InQuirY 3
In groups
Aim: To know surds
CHAPTER 4
Instructions:
1. Observe the diagram on the right.
2. Without using a calculator, find the value of cosine θ and give
the answer in the form of a , where a and b are integers.
!wb
3. Discuss your group's findings.
3 cm
2 cm
We often face problems as mentioned above. How can problems involving surds be solved?
Let's explore.
Comparing rational numbers and irrational numbers, and relating
surds with irrational numbers
You have learnt about rational numbers, in other words, numbers that can be expressed in
a
the form of , where a and b are integers and b ≠ 0. Rational numbers can also be written in
b
1
decimal form such as = 0.3333… What is the connection between rational numbers and
3
irrational numbers?
InQuirY 4
In groups
21st Century Learning
Aim: To find the relation between surds and irrational numbers
Instructions:
1. Scan the QR code or visit the link on the right.
2. Browse the Internet to obtain information about surds.
3. Cut all the prepared number cards and paste them onto the table
according to the correct classification as shown as the following.
Rational numbers
bit.ly/315G1Ub
Irrational numbers
Surd
Not surd
0.333333…
4. Convert all decimals on the number cards to fractions. What can be concluded?
5. Each group will move to other groups to see their findings.
6. Discuss with team members about the findings of other groups.
96
4.2.1
Indices, Surds and Logarithms
(a) Decimals that can be coverted to fractions are rational
numbers.
(b) Decimals that cannot be converted to fractions are
irrational numbers.
(c) Numbers with radicals, if the values are integers or a
recurring decimals, are not surds.
a , where
Surds are numbers in the square root form, that is !w
a is any positive integer. Surds have infinite decimal places and
a is called "surd a order n". For example, 3!w
4
are non-recurring. n!w
is called "surd 4 order 3". When a number cannot be simplified by
eliminating the root, then that number is classified as a surd.
MATHEMATICS
POCKET
• Radical symbols are as
follows.
!w, 3!w, 5!w, n!w
• Recurring decimals
are decimals that
can be converted to
fractions. An example
of a recurring decimal is
54.5656…
CHAPTER 4
From the results of Inquiry 4, it is found that:
For instance,
2 cannot be simplified, therefore !w
2 is a surd.
(a) !w
4 can be simplified as 2, therefore !w
4 is not a surd.
(b) !w
Are all numbers with roots, surds? Observe the following table.
Mathematics Museum
Number
Simplified
number
Decimal
Surd or
Not surd
3
!w
3
!w
1.7320508...
Surd
!w4l
1
2
0.5
Not surd
11
!w
2.2239800...
Surd
3
3
Not surd
3
!w
1.2457309...
Surd
11
!w
3
27
!w
3
3
!w
5
3
5
1
1
4 3 2
5
1 1
6
7
8
9
10 11 12 13
17
16
15
14
In geometry, the first
Theodorus spiral (also called
square root spiral, Einstein
spiral or Pythagorean spiral)
was built by Theodorus from
Cyrene. This spiral composed
of right-angled triangles that
were placed edge-to-edge.
From the table above, it is found that surds have non-recurring
decimals. Therefore, surds are irrational numbers. Recurring
decimals, such as, 54.565656… are sometimes written as 54.56 or
54.56
.
4.2.1
97
Example 6
Convert the following recurring decimals to fractions.
(a) 0.676767…
(b) 12.645645645…
Solution
CHAPTER 4
(a) Let,
2 –
N = 0.676767… … 1
100N = 67.6767… … 2
1 : 99N = 67
N = 67
99
Thus, 0.676767… = 67.
99
(b) Let,
A = 12.645645645…
A = 12 + N
Assume, N = 0.645645645… …
…
1000N = 645.645645…
2 – 1 : 999N = 645
N = 645
999
= 215
333
A = 12 + 215
333
Thus, 12.645645645… = 12215.
333
Multiply with a
suitable integer so
that the part with the
recurring decimal can
be eliminated.
Mind Challenge
Convert the following
fraction to a recurring
decimal.
224
495
1
Smart TIPS
2
Is !w
a = n!w
a?
n
1
!w9 = (9)2 = 3
1
9 = 2 × 92
2!w
=2×3
=6
Since 3 ≠ 6, thus
n
!wa ≠ n!wa .
Example 7
Determine whether the following terms are surds. Give your reason.
125
(a) 3!w
(b) 5!w
125(c)
!16
64
4
Solution
Use a scientific calculator to obtain the values.
1
(a) 3!w
125 = 125 3
= 5
3
125 is not a surd because the value is an integer.
!w
(b) 5!w
125 = 2.6265278
5
125 is a surd because it is a non-recurring decimal.
!w
98
4.2.1
Indices, Surds and Logarithms
(c)
= 0.7071067…
!16
64
is a surd because it is a non-recurring decimal
!16
64
4
4
Example 8
Smart TIPS
Is !w
4 = 2!w
4 ? Explain.
Solution
1
!wa ≠ n!wa because n!wa = an
1
a = n × a2.
while n!w
n
1
1
CHAPTER 4
4 = 4 2 , 2!w
4 = 2 × 42
!w
=2
=2×2
=4
Since 2 ≠ 4, thus !w
4 ≠ 2!w
4. Generally, n!w
a ≠ n!w
a.
Self Practice 4.3
1. Convert the following recurring decimals to fractions.
(a) 0.787878…
(b) 3.57575757…
(c) 0.345345345…
(d) 13.567567567...
2. Determine whether the following terms are surds. Give your reason.
(a) 3!w
127
(b) 4!w
1 125
(c) 6 64
(d)
729
!
79
!897
7
Making and verifying conjectures on !w
a  !w
b and !w
a  !w
b
InQuirY 5
In groups
Aim: To verify conjectures on !wa × !wb and !wa ÷ !wb
Instructions:
1. Scan the QR code or visit the link on the right.
bit.ly/2MkBDv5
2. Click on the "Law 1" and "Law 2" boxes. Then, drag cursors a and b.
3. State the conjectures based on your observation regarding both laws.
4. With a scientific calculator, fill in the following tables by taking any positive integer a
and b.
4.2.1
a
b
(a  b)
!wa
!wb
Value of !wa  !wb
(a  b)
!w
Value of !w
(a  b)
2
5
10
!w2
!w5
3.162…
10
!w
3.162…
4.2.2
99
a
b
(a  b)
!wa
!wb
Value of !wa  !wb
(a  b)
!w
Value of !w
(a  b)
10
5
2
10
!w
!w5
1.414…
!w2
1.414…
5. Compare the values of the 6th row and 8th row for both completed tables.
CHAPTER 4
6. Were you able to verify the conjectures made? Discuss.
From the results of Inquiry 5, it is found that:
For a > 0 and b > 0,
a × !w
b = !w
ab
(a) !w
(b) !w
a ÷ !w
b= a
b
!
MATHEMATICS
POCKET
(Law 1)
If a  0, then !w
a is a real
number and
(Law 2)
a2 = (a2)2 = a
!wa × !wa = !w
1
Example 9
Write the following as single surds.
2 × !w
7
(a) !w
(b)
(c) !w
3a × !w
5a
(d)
Solution
100
(b)
!
21a
!w
= 21a
7a
7a
!w
15a2
= !w
= !w
3
15
= a!w
3a × 5a
(c) !w
5a = !w
3a × !w
21a
!w
7a
!w
24
!w
= 24
8
8
!w
14
= !w
= !w
3
(a) !w
2 × !w
7 = !w
2×7
24
!w
8
!w
(d)
!
4.2.2
Indices, Surds and Logarithms
Self Practice 4.4
1. Write the following as single surds.
2 × !w
3
(b) !w
3 × !w
5
(a) !w
(e)
8
!w
3
!w
(f)
18
!w
3
!w
(c) !w
3 × !w
3
(d) !w
5 × !w
6
(g)
(h)
20
!w
5
!w
5 × !w
6
!w
3
!w
Simplifying expressions involving surds
Individual
Aim: To simplify expressions involving surds
Instructions:
1. Scan the QR code or visit the link on the right.
2. Drag the cursor to change the value of the surd.
bit.ly/2LQqH9u
3. Write down the surds that can be simplified and the surds
that cannot be simplified.
90 without using mathematical tools and technological devices.
4. Simplify !w
3 cannot be simplified but !w
9 can be simplified
From the results of Inquiry 6, it is found that !w
90 can be written as !w
9 × 10 or !w
9 × !w
10, therefore !w
90 = 3!w
10.
to 3. Besides, !w
Example 10
Write !w
18 in the form of a!w
b where a and b are integers and a is the largest value.
Solution
18 = !w
9×2
!w
9 × !w
2
= !w
2
= 3!w
9 is the largest perfect square number
and a factor of 18
Self Practice 4.5
1. Mark (✓) on the correct statements.
5 !w
7
!w
= !w
12
4!w
8
4
2!w
2
= 2!w
4.2.2
4.2.3
= 6!w
2
260
!w
= 2!w
65
(!w
16!w
36 )
4!w
7 × 5!w
7
= 576
= 20!w
21
18
!w
3
!w
15
= !w
75
!w
3
!w
=5
30!w
27
3
6!w
= 15
(!w
81 )
2 × 2!w
2
3!w
2
2
= 81
101
CHAPTER 4
InQuirY 6
b where a and b are integers and a is the largest value.
2. Write the following in the form of a!w
(a) !w
12
(e) !w
45
(b) !w
27
(f) !w
48
(c) !w
28
(g) !w
54
(d) !w
32
(h) !w
108
How to perform addition, subtraction and multiplication involving surds? Let's explore this in
detail.
CHAPTER 4
InQuirY 7
In groups
21st Century Learning
Aim: To perform mathematical operations involving additions,
subtractions and multiplications on surds
Instructions:
1. Scan the QR code or visit the link on the right.
bit.ly/31OY9To
2. Consider the expressions involving surds.
3. Click on the "Solution" button to see the calculation steps.
4. Click on "Other Questions" to see the next question.
5. Make a note about the calculation steps shown and explain to other classmates about
your understanding on solving expressions involving surds.
From the results of Inquiry 7, it is found that:
Expressions involving surds can be simplified by performing addition, subtraction and
multiplication operations of surds.
Example 11
Simplify the following expressions.
2 × !w
3 + !w
6
(a) !w
(c) !w
18 – !w
8
(b) !w
7 (6 – !w
7)
(d) (6 + 2!w
2 )(1 + 3!w
2)
Solution
(a) !w
2 × !w
3 + !w
6 = !w
2 × 3 + !w
6
(b) !w
7 (6 – !w
7 ) = 6!w
7 – !w
7 × !w
7
6 + !w
6 = 6!w
7–7
= !w
6
= 2!w
(c) !w
18 – !w
8
(d) (6 + 2!w
2 )(1 + 3!w
2)
= !w
9 × 2 – !w
4 × 2 = !w
9 × !w
2 – !w
4 × !w
2
2 – 2!w
2 = 3!w
2
= (3 – 2)!w
2
= !w
102
= 6(1) + 6(3!w
2 ) + 2!w
2 (1) + (2!w
2 )(3!w
2)
= 6 + 18!w
2 + 2!w
2 + 12
= 18 + 20!w
2
4.2.3
Indices, Surds and Logarithms
Example 12
b.
Simplify each of the following in the form of a!w
(a) 4!w
27
(b) 7!w
243
(c) 5!w
75
(b) 7!w
243 = 7!w
81 × 3
(c) 5!w
75 = 5!w
25 × 3
Solution
= 4(3)!w
3 = 12!w
3 = 7(9)!w
3 = 63!w
3 = 5(5)!w
3
= 25!w
3
In Example 12, notice that 12!w
3, 63!w
3 and 25!w
3 have !w
3 as the irrational factor. Therefore,
these three expressions are known as similar surds.
Numbers that do not contain a similar irrational factor are known as not similar surd. For
3 , 23!w
3, 5!w
6 and 74!w
3 are not similar surds.
example, the set of expressions !w
Example 13
Determine whether the set of expressions 4!w
12 , 5!w
18 and 5!w
6 are similar surds or not similar
surds.
Solution
4!w
12 = 4!w
4×3
5!w
18 = 5!w
9×2
= 4(2)!w
3 = 8!w
3 5!w
6 = 5!w
2×3
= 5(3)!w
2 = 15!w
2
= 5!w
6
The three expressions do not contain a similar irrational number factor. Thus, all of those
expressions are not similar surds.
Self Practice 4.6
1. Simplify the following expressions involving surds.
(a) 3!w
5 + 5!w
5
(d) !w
6 (3!w
6 – 5!w
6)
(g) (4 + 5!w
3 )(3 + 5!w
3)
(b) 7!w
5 + 5!w
5
(e) !w
5 (4 + 5!w
5)
(h) (7 – 5!w
7 )(3 + 5!w
7)
(c) 7!w
7 – 5!w
7
(f) !w
7 (3 – 5!w
7)
(i) (9 + 5!w
4 )(3 – 5!w
4)
2. Determine whether the following sets of expressions are similar surds or not similar surds.
(a) 5!w
80 , 2!w
58 , 9!w
45
(b) 3!w
3 , 4!w
12 , 5!w
27
(c) 2!w
125, 7!w
5 , –7!w
5
(d) 2!w
12 , 9!w
24 , 8!w
5
(e) 3!w
27 , –3!w
27 , –!w
3
4.2.3
103
CHAPTER 4
(a) 4!w
27 = 4!w
9×3
Rationalising the denominators for expressions involving surds
CHAPTER 4
1
1
and
,
Numbers that contain irrational denominators such as 1 ,
a m!w
a + n!w
b
a – n!w
b
m!w
m!w
where m and n are integers should be written by rationalising the denominators. The rules to
rationalising the denominators are as follows:
(a) Multiply the numerator and denominator of 1 with the conjugate surd m!w
a so that the
a
m!w
surd can be eliminated from the denominator.
1
with the conjugate surd m!w
(b) Multiply the numerator and denominator of
a – n!w
b
a + n!w
b
m!w
so that the surd can be eliminated from the denominator.
1
with the conjugate surd m!w
(c) Multiply the numerator and denominator of
a + n!w
b
a – n!w
b
m!w
so that the surd can be eliminated from the denominator.
Example 14
Rationalise the denominator and simplify each of the following.
1
1
(c)
(a) 1 (b)
3
2 + 5!w
3
3 – 5!w
7
5!w
7!w
2!w
Solution
5!w
3
(a) 1 = 1 ×
3 5!w
3 5!w
5!w
3
Multiply with the conjugate surd
5!w
3
3 × !w
3
5 × 5 × !w
5!w
3
=
75
3
!w
=
15
MATHEMATICS
POCKET
Rationalising using
conjugate surds.
Conjugate
surd
Surd
m!wa
m!wa
m!wa + n!wb
m!wa – n!wb
m!wa – n!wb
m!wa + n!wb
=
7!w
2 – 5!w
3
1
1
=
×
2 + 5!w
3 7!w
2 + 5!w
3 7!w
7!w
2 – 5!w
3
7!w
2 – 5!w
3
=
(7!w2 + 5!w3 )(7!w2 – 5!w3 )
7!w
2 – 5!w
3
=
2
(7!w2 ) – (5!w3 )2
7!w
2 – 5!w
3
=
23
(b)
104
Multiply with the
conjugate surd
Smart TIPS
!wa × !wa = (!wa ) = a
(a − !wb )(a + !wb ) = a2 − b
2
4.2.4
Indices, Surds and Logarithms
2!w
3 + 5!w
7
1
1
=
×
3 – 5!w
7 2!w
3 – 5!w
7 2!w
2!w
3 + 5!w
7
2!w
3 + 5!w
7
=
(2!w3 – 5!w7 )(2!w3 + 5!w7 )
2!w
3 + 5!w
7
=
2
(2!w3 ) – (5!w7 )2
2!w
3 + 5!w
7
=–
163
Multiply with the
conjugate surd
(c)
The conjugate surd for
2!w
3 – 5!w
7 is 2!w
3 + 5!w
7.
Rationalise the denominator and simplify
1 + !w
3
.
3
1 – !w
Solution
1 + !w
3 1 + !w
3 1 + !w
3
=
×
3 1 – !w
3 1 + !w
3
1 – !w
1 + 3 + !w
3 + !w
3
=
1–3
4 + 2!w
3
=
–2
3
= –2 – !w
Mind Challenge
Multiply with the
conjugate surd
What is the conjugate surd
of 1 – !w3?
Smart TIPS
a – !wb
with a
a – !wb
c
fraction in the form
a + !wb
to eliminate the surd from
the denominator.
Multiply
Example 16
Write
5 + !w
7 4 – !w
7
as a single fraction.
+
3 1 – !w
3
1 + !w
Solution
(
)(
)
5 + !w
7 4 – !w
7
5 + !w
7 1 – !w
3
4 – !w
7 1 + !w
3
=
×
×
+
+
3 1 – !w
3
3 1 – !w
3
3 1 + !w
3
1 + !w
1 + !w
1 – !w
5 – 5!w
3 + !w
7 – !w
21 + 4 + 4!w
3 – !w
7 – !w
21
=
(1 + !w3 )(1 – !w3 )
9 – !w
3 – 2!w
21
=
1–3
–9 + !w
3 + 2!w
21
=
2
4.2.4
BRAINSTORMING
“The product of 2 irrational
numbers is an irrational
number.”
Discuss and give your
justification regarding this
statement.
105
CHAPTER 4
Example 15
Self Practice 4.7
1. Rationalise the denominator and simplify each of the following:
(a) 2
5
!w
3
!w
(d)
12
2!w
6 – !w
3
(g)
12
9 – !w
(b) 7
2
!w
1 + !w
3
(e)
12
!w
3 + !w
2 4 – !w
3
(h)
+
2 7 + !w
3
5 – !w
2
!w
5
!w
3 + !w
2
(f)
5
5 – !w
7 – !w
5 6 + !w
3
(i)
–
5 6 – !w
3
5 + !w
(c)
CHAPTER 4
Solving problems involving surds
Example 17
Mathematics
Application
The diagram on the right shows a pyramid-shaped house. The
triangle shape at the front of the house has an area of
(20!w3 – 4) m
2
and the length of its base is (4 + 4!w
3 ) m.
Determine the height of the triangle at the front of the house in
the form of (a + b!w
3 ), where a and b are rational numbers.
Solution
1 . Understanding the problem
◆
◆
◆
The area of the triangle = (20!w
3 – 4) m2
The length of the base of the triangle
= (4 + 4!w
3) m
Determine the height of the triangle in
the form of (a + b!w
3)
2 . Planning a strategy
◆
Use the formula of the area of triangle
1
= × base × height
2
3 . Implementing the strategy
1
× (4 + 4!w
3 ) × t = 20!w
3–4
2
(2 + 2!w3 )t = 20!w3 – 4
20!w
3–4
3
2 + 2!w
20!w
3 – 4 2 – 2!w
3
=
×
3
3
2 + 2!w
2 – 2!w
t=
40!w
3 – 120 – 8 + 8!w
3
–8
–128 + 48!w
3
=
–8
3
= 16 – 6!w
=
The height of the triangle part at the front
of the house is (16 – 6!w
3 ) m.
106
4.2.4
4.2.5
Indices, Surds and Logarithms
4 . Making a conclusion
1
× (4 + 4!w
3 ) × (16 – 6!w
3)
2
= (2 + 2!w
3 )(16 – 6!w
3)
The area of triangle =
= 32 – 12!w
3 + 32!w
3 – 36
= (20!w
3 – 4) m2
Example 18
CHAPTER 4
Solve x – 4!w
x + 3 = 0.
Solution
x – 4!w
x +3=0
Factorise
(!wx – 3)(!wx – 1) = 0
x – 3 = 0 or !w
x –1=0
!w
x=3
x =1
!w
!w
2
2
(!wx ) = 3
(!wx )2 = 12
x=9
x =1
Self Practice 4.8
1. Triangle ABC has an angle of ABC = 60°, AB = 3!w
3 cm and BC = 4!w
3 cm. Determine the
length of AC.
A
2. The diagram on the right shows a right-angled triangle ABC.
(a) Determine the area of triangle ABC.
2 ) cm
(5 – 2
(b) Determine the length of AC.
B
2 ) cm
(5 + 2
C
y = 6!w
3 + 5. Write your answer in the form of a + b!w
3 , where a and
3. Solve the equation 2 + 3!w
b are rational numbers.
4. Solve the following equations.
2 – 7x + 2x = 0
(a) !w
2x + 1 + !w
2x – 1 = 2
(b) !w
4x + 3 – !w
4x – 1 = 2
(c) !w
4.2.5
107
Intensive Practice 4.2
Scan the QR code or visit bit.ly/33f32pd for the quiz
1. Write the following as single surds.
5 × !w
11
(a) !w
(b) !w
7 × !w
10
(c)
27
!w
18
!w
(d)
48
!w
8
!w
2. Write the following in the form of a!w
b, where a and b are integers and a is the largest value.
24
(a) !w
CHAPTER 4
3. Simplify.
(b) !w
162
(c)
54
!w
3
!w
(d)
( )
2!w
6
3
2
(a) 3!w
10 + 5!w
10
(b) 6!w
11 – !w
11
(g) 3!w
15 × 7!w
5
(h) !w
72 × 4!w
15
(i) !w
4 (2!w
3 ) – 5!w
3
(m) (7 + 5!w
7 )(3 – 5!w
7)
(n) (7 – 5!w
5 )(3 – 5!w
5)
(o)
(d) 2!w
45 + !w
20
(j) !w
7 (3 + 7!w
7)
(p)
12
!w
108
!w
(e) 3!w
27 – !w
72
(k) !w
5 (7 – 5!w
5)
(q)
88
!w
11
2!w
(c) 13!w
13 – 2!w
13
(f) !w
18 + !w
27
(l) (3 + 3!w
7 )(3 + 5!w
7)
112
!w
7
!w
9!w
20
(r)
5
3!w
4. Given A = 3!w
5 + 7!w
3 , B = 2!w
5 – 7!w
7 and C = 2!w
3 – 9!w
8 . Simplify
(a) A + B
(b) A – C
(c) 3A + 2B
(d) 3A + B – 2C
5. Rationalise the denominators and simplify the following expressions.
4
4
(b)
(c)
(a) 2
5
5
5
3
–
3
–
3!w
!w
!w
4 + !w
5
3 – !w
7
!w
5
(d)
(e)
(f)
3 – !w
2
5
3 + !w
7
2!w
3 – !w
!w
6. Write the following as single fractions.
1 + 1
2
1
(a)
(b)
+
3 1 – !w
3
7 + !w
2 !w
7 – !w
2
1 + !w
!w
(c)
2 + 1
3 4 + !w
3
4 – !w
7. The area of a rectangle is (8 + !w
10 ) cm2. One of its sides has a length of (!w
5 + !w
2 ) cm.
Determine the length of the other right in the form of a!w
5 + b!w
2.
8. The diagram on the right shows a right-angled triangle PQR.
(a) Determine the value of tan x. Write your answer in the
a + b!w
2
form of
, where a, b and c are integers.
2 ) cm
(3 + c
(b) Determine the area of triangle PQR. Write your
p + q!w
2
Q
answer in the form of
, where p, q and r
r
are integers.
108
P
2 ) cm
(1 + 2 x
R
Indices, Surds and Logarithms
4.3
Laws of Logarithms
Relating the equations in index form with logarithmic form and
determining the logarithmic value of a number
If am = an then, m = n
If am = bm then, a = b
CHAPTER 4
An equation in index form can be written as N = ax where a > 0
and a ≠ 1. N, a and x are variables. We can find the value of one
variable if the value of the other two variables are given.
For example,
(a) if 81 = 9x, then x = 2
1 000
(b) if 1 000 = a3, then a = 3!w
= 10
(c) if N = 53, then N = 125
Can you find the value of x of the following equations?
(a) 50 = 4x
(b) 69 = 7x
(c) 80 = 8x
What is the method that can be used? Let's explore in detail.
Inquiry 8 will explain the methods to solve the equation above.
InQuirY 8
In groups
Aim: To relate the equations in index form and logarithmic form
Instructions:
1. Scan the QR code or visit the link on the right.
bit.ly/30Txh3t
2. Click on the "Graph of equation in index form" box and
x
observe the graph f(x) = a that appears.
3. Then, click on the "Graph of equation in logarithmic form" box and observe the graph
of g(x) = loga(x) that appears.
4. Drag cursor a to the left and to the right. Write down your observations regarding the
changes that happen to the graph when the value of a increases.
5. Drag cursor a to the value of 1. Does the graph g(x) = loga x exist? What is the shape of
the graph for f (x) = ax that was formed? Write down your findings.
6. Drag cursor a to negative values. Does the graph f (x) = ax and g(x) = loga x exist? Write
down your findings.
7. Discuss on the existence of logarithms for negative numbers and zero.
8. Then, verify if the following statements are true or false.
(b) loga a = 1
(a) loga 1 = 0
4.3.1
109
Smart TIPS
From the results of Inquiry 8, it was found that the relationship
between equations in index and logarithmic forms can be defined
as follows:
loga ax = x
loga N = x ⇔ N = ax where a > 0 and a ≠ 1
From the definition above, it can be concluded that:
a0 = 1 ⇔ loga 1 = 0
and
a1 = a ⇔ loga a = 1
CHAPTER 4
Thus, for any real numbers, a > 0 and a ≠ 1, the following
statement is true.
loga 1 = 0
loga a = 1
MATHEMATICS
POCKET
Index
form
Logarithmic
form
40 = 1
log4 1 = 0
100 = 1
log10 1 = 0
71 = 7
log7 7 = 1
10 = 10
log10 10 = 1
1
Notice that:
loga N is defined if N > 0 and a > 0, a ≠ 1
For example, log7 0, log10 (−10), log0 2 and log1 13 are undefined.
The base of logarithms must be positive. Usually, 1 is not used
as a base because 1n = 1 for any value of n.
If a common logarithm value is given for a number, that
number can be determined with a scientific calculator. That
number is called antilogarithm or antilog for short.
MATHEMATICS
POCKET
The common logarithm is
the logarithm in base 10.
For example, log10 a = lg a
If log10 N = x, then antilog x = N
Based on the definition of logarithms of a number, we can convert
an index equation to the logarithmic form.
The power of the index number is the logarithmic value
Given 16 = 24 thus log2 16 = 4
The base of the index number is the base of the logarithm
Mathematics Museum
The value of common
logarithms can be
determined by using a
scientific calculator or a
four-figure table.
Scan the QR code below to
obtain a four-figure table.
Otherwise, we can convert an equation in logarithmic form to
index form.
If log2 16 = 4, then 16 = 2
bit.ly/2MvAcvb
4
110
4.3.1
Indices, Surds and Logarithms
InQuirY 9
In pairs
21st Century Learning
Aim: To relate the graphs of exponential function and logarithmic function.
Instructions:
1. Copy and complete the table below for y = 2x.
x
–3
y
1
8
–2
–1
0
1
2
3
x
1
8
y
–3
CHAPTER 4
2. Then, copy and complete the table below for the inverse function y = 2x, by exchanging
the value of x and y.
3. Draw the graph of y against x for y = 2x and its inverse function on the same axis.
4. Note down your observation about both of the graphs drawn.
5. Present your findings in front of the class.
From Inquiry 9, f : x → 2x, x = f −1(2x).
Suppose y = 2x,
thus x = f −1(y)
log2 y = log2 2x
log2 y = x
Substitute x = log2 y into x = f −1(y)
then, f −1(y) = log2 y
or f −1(x) = log2 x
Generally,
y
y = ax
y=x
(0, 1)
0
y = loga x
x
(1, 0)
If f : x → ax, then f −1 : x → loga x
Thus,
y = loga x is the inverse of ay = x
Example 19
Convert 24 = 16 to logarithmic form.
Solution
24 = 16
log2 16 = 4
4.3.1
111
Example 20
Convert log3 27 = 3 to index form.
Solution
log3 27 = 3
33 = 27
Example 21
Find the value of each of the following.
CHAPTER 4
(a) log10 7
()
(c) log10 3
4
(b) log10 79
Solution
()
3
( )
3
(c) log10 3 = log10 27
4
64
= − 0.3748
(a) log10 7 = 0.8451
(b) log10 79 = 1.8976
Example 22
Find the value of each of the following.
(a) log5 625
(b) log6 7 776
Solution
(a) Let, log5 625 = x
Let, log6 7 776 = y
x
6y = 7 776
5 = 625
x
4
6y = 65
5 = 5 x = 4
y =5
Thus, log5 625 = 4 Thus, log6 7 776 = 5
Example 23
(a) Determine the value of x if log5 x = 3.
(b) Determine the value of y if log3 y = 4.
Solution
(a) log5 x = 3
(b) log3 y = 4
3
x = 5 y = 34
x = 125 y = 81
112
4.3.1
Indices, Surds and Logarithms
Example 24
Determine the value of each of the following.
(a) antilog 0.1456
(b) antilog (− 0.3976)
Solution
(a) antilog 0.1456 = 1.3983
(b) antilog (− 0.3976) = 0.4003
Self Practice 4.9
2. Convert the following to index form.
(a) log10 10 000 = 4
(c) log2 128 = 7
3. Find the value of each of the following:
(a) log10 9
3
(c) log10 5
6
(e) log3 81
(g) log10 100 000
()
(c) 53 = 125
(d) 63 = 216
CHAPTER 4
1. Convert the following to logarithmic form.
(b) 27 = 128
(a) 34 = 81
(b) log10 0.0001 = −4
(d) log4 64 = 3
(b) log10 99
(d) log2 64
(f) log4 256
4. Solve the following equations.
(a) log2 x = 5
(b) log8 x = 3
(c) log2 x = 8
5. Determine the value of each of the following:
(a) antilog 2.1423
(b) antilog 1.3923
(c) antilog 3.7457
(d) antilog (−3.3923)
(e) antilog (−2.5676)
(f) antilog (− 4.5555)
Proving the laws of logarithm
InQuirY 10
In groups
21st Century Learning
Aim: To prove the laws of logarithm
Instructions:
1. Scan the QR code or visit the link on the right.
2. Observe the examples of three laws of logarithm that is shown.
3. Drag cursors a, b and n. Observe the changes to the three laws of
logarithm.
4. Discuss the three laws of logarithms and make a conclusion.
5. Make a short presentation about your findings.
4.3.1
4.3.2
bit.ly/2MiX9QP
113
From the results of Inquiry 10, three basic laws of logarithm are as follows:
If a, x and y are positive and a ≠ 1, then
(a) loga xy = loga x + loga y
(b) loga x = loga x − loga y
y
(c) loga xn = n loga x for any real number n
(Product law)
(Division law)
(Power law)
Each basic law of logarithm above can be proven as follows:
Let x = ap and y = aq, then p = loga x and q = loga y.
CHAPTER 4
(a) xy = ap × aq = ap + q
Thus, loga xy = p + q
loga xy = loga x + loga y
p
(b) x = aq = ap − q
y a
Thus, loga x = p – q
y
loga x = loga x − loga y
y
(c) xn = (ap)n = apn
Thus, loga xn = pn
loga xn = n loga x
From the definition of logarithms
Substitute p = loga x and q = loga y
From the definition of logarithms
Substitute p = loga x and q = loga y
From the definition of logarithms
Substitute p = loga x
Example 25
Given log5 15 = 1.6826 and log5 4 = 0.8614. Determine the
value of each of the following.
(a) log5 60 (b) log5 12 (c) log5 100
Solution
(a) log5 60 = log5 (15 × 4)
= log5 15 + log5 4
= 1.6826 + 0.8614
= 2.544
(b) log5 12 = log5 60
5
= log5 60 − log5 5
= 2.544 – 1
= 1.544
(c) log5 100 = log5 (25 × 4)
= log5 25 + log5 4
= log5 52 + log5 4
= 2 log5 5 + 0.8614
= 2 + 0.8614
= 2.861
( )
114
Tech Whizz
loga ax = x
Check your answers with
the Photomath app. Scan
the QR code below to
download the Photomath
app.
bit.ly/2Rg86YH
4.3.2
Indices, Surds and Logarithms
Example 26
Find the value of each of the following without using a calculator.
(b) log3 8 + 2 log3 6 − log3 96
9
(a) log5 750 − log5 6
Solution
(a) log5 750 − log5 6 = log5 750
6
= log5 125
= log5 53
loga ax = x
= 3 log5 5
=3
(b) log3 8 + 2 log3 6 − log3 96 = log3 8 + log3 62 − log3 96
9
9
= log3 8 × 36 ÷ 96
9
= log3 27
= log3 33
loga ax = x
= 3 log3 3
=3
)
CHAPTER 4
(
Self Practice 4.10
1. Given that log7 4 = 0.712 and log7 5 = 0.827. Evaluate each of the following:
(a) log7 1 1
(c) log7 100
(d) log7 0.25
(b) log7 28
4
2. Evaluate each of the following without using a calculator.
(a) log3 21 + log3 18 – log3 14
(b) 2 log4 2 – 1 log4 9 + log4 12
2
(c) log2 7 + log2 12 – log2 21
Simplifying algebraic expressions using the laws of logarithms
Algebraic expressions involving logarithm can be simplified using the laws of logarithms.
Example 27
Express the following as single logarithms.
(a) loga x + 3 loga y
(b) 2 loga x – 1 loga y
2
(c) 2 log3 x + log3 y – 1
Solution
(a) loga x + 3 loga y = loga x + loga y3
= loga xy3
4.3.2
4.3.3
115
1
(b) 2 loga x – 1 loga y = loga x2 – loga y 2
2
2
= loga x
y
!w
(c) 2 log3 x + log3 y – 1 = log3 x2 + log3 y – log3 3
x2y
= log3
3
CHAPTER 4
Example 28
If p = logb 2, q = logb 3 and r = logb 5, write the following in
terms of p, q and/or r.
(b) logb 45
(a) logb 6
5!w
3
(d) logb
(c) logb 0.2222…
2
( )
Solution
(a) logb 6 = logb (2 × 3)
= logb 2 + logb 3
= p + q
(b) logb 45 = logb (9 × 5)
= logb 32 + logb 5
= 2 logb 3 + logb 5
= 2q + r
(c) logb 0.2222… = logb 2
9
= logb 2 − logb 9
= logb 2 − logb 32
= logb 2 – 2 logb 3
= p – 2q
( )
5!w
3
= logb 5 + logb !w
3 − logb 2
2
= logb 5 + 1 logb 3 − logb 2
2
1
=r+ q–p
2
(d) logb
Mind Challenge
Can you determine the value
of
(a) log10 (–6)?
(b) log–10 6?
FLASHBACK
Suppose,
A = 0.2222…
100A = 22.22…
1
2
2 – 1 : 99A = 22
22
99
2
=
9
A=
Self Practice 4.11
1. Write the following expressions as single logarithms.
(b) logb x – 3 logb y
(c) log2 x + 3 log2 y
(a) log2 x + log2 y2
1
4
(d) log4 x + 2 – 3 log4 y
(e) log3 m + 2 log3 n – log3 m
2
2. Given log2 3 = p and log2 5 = q, express each of the following in terms of p and q.
(b) log2 45
(c) log2 !w
15
(a) log2 10
116
4.3.3
Indices, Surds and Logarithms
logarithm of a number
logc b
and determining the
logc a
If a, b and c are positive numbers, then, a ≠ 1 and c ≠ 1,
logc b
then loga b =
logc a
The prove for the above statement are as follows:
Suppose loga b = x, then, ax = b.
Use the same logarithmic
logc ax = logc b
base on both sides
x logc a = logc b
logc b
Power law of logarithm
x=
logc a
logc b
Thus, loga b =
logc a
QR
Logarithmic base changes.
bit.ly/2NtoekS
CHAPTER 4
Proving the relationship of loga b =
Specifically:
If b = c, then loga b =
logb b
1
=
logb a logb a
By using the law of base change, any logarithmic base can be
written and evaluated with base 10 or base e.
Logarithms with base e are known as natural logarithms
and are written as loge, or ln. Base e is often used in mathematics,
science and technology.
Example 29
Determine the following values by changing their bases to 10.
(b) log2 0.45
(a) log30 4
Solution
log10 4
log10 0.45
(b) log2 0.45 =
log10 30
log10 2
= – 0.3468
= 0.6021
1.4771
0.3010
= 0.408
= –1.152
MATHEMATICS
POCKET
ln a means loge a with e
as an exponent base. The
number e is a non-recurring
decimal, which is 2.7182…
Observe the following:
• log 10 = 1
• ln e = 1
• ln ex = x
• eln x = x
• 10log x = x
(a) log30 4 =
4.3.4
Mind Challenge
Find the value of log5 20
using common logarithms
and natural logarithms.
117
Example 30
Convert each of the following to natural logarithms and
evaluate them.
(b) log30 4
(a) log6 254
Solution
CHAPTER 4
Determine the solution to
Example 30 with a scientific
calculator.
1. Press In 254 ) ÷
loge 254
loge 4
(b) log30 4 =
loge 6
loge 30
ln
254
= ln 4
=
ln 6
ln 30
= 5.5373 = 1.3863
1.7918
3.4012
= 3.090
= 0.408
(a) log6 254 =
Quick Count
In 6 ) = .
2. The screen will show:
In(254) ÷ In(6)
3.090445097
Example 31
Given log5 x = p, express each of the following in terms of p.
(b) logx 25x3
(a) log25 x
Solution
log5 x
log5 25x3
(b) logx 25x3 =
log5 x
log5 25
p
log5 52 + log5 x3
= 2 =
p
2 log5 5 + 3 log5 x
=
p
2 + 3p
=
p
(a) log25 x =
Self Practice 4.12
1. Evaluate each of the following by converting it to base 10.
(b) log6 1.32
(c) log5 18
(a) log3 22
(d) log4 0.815
2. Convert each of the following to natural logarithms and evaluate them.
(a) log7 225
(b) log9 324
(c) log20 379
3. Given log3 2 = t, express each of the following in terms of t.
(a) log2 9
(b) log9 8
(c) log2 18
(d) log2 9
4
4. If log2 m = a and log2 n = b, express each of the following in terms of a and b.
(a) log4 m2n3
(b) log8 m2
(c) logmn 8n
n
118
4.3.4
Indices, Surds and Logarithms
Solving problems involving the laws of logarithms
Problems involving indices, such as 3x = 70 that cannot be expressed in the form of ax = ay or
ax = bx can be solved by using logarithms.
Example 32
Solve the equation 3x − 4 = 50x − 3.
Solution
Use base 10 logarithm
log10 a = log a
CHAPTER 4
3x − 4 = 50x − 3
(x − 4) log 3 = (x − 3) log 50
x log 3 – 4 log 3 = x log 50 – 3 log 50
x log 3 – x log 50 = –3 log 50 + 4 log 3
x (log 3 – log 50) = −3 log 50 + 4 log 3
–3 log 50 + 4 log 3
x=
log 3 – log 50
= 2.610
Example 33
Solve the following natural logarithmic equations.
(b) 10e2x = 35
(a) ln (4x − 2) = 5
Solution
(a) ln (4x − 2) = 5
(b) 10e2x = 35
loge (4x − 2) = 5 e2x = 3.5
e5 = 4x − 2 ln e2x = ln 3.5
148.4132 = 4x − 2 2x ln e = ln 3.5
4x = 150. 4132 2x = ln 3.5
x = 150.4132 x = ln 3.5
4
2
= 37.603= 0.626
Example 34
Mathematics
ln e = 1
Application
The temperature of a block of steel rises from 30°C to T °C when it was heated for x seconds.
Given T = 30(1.2)x, determine
(a) the temperature of the steel when it is heated for 10.4 seconds,
(b) the time, x, in seconds, taken to increase the temperature of the block of steel from 30°C
to 1 500°C.
4.3.5
119
Solution
1 . Understanding the problem
Given the formula T = 30(1.2)x
The temperature rises from 30°C to
T °C.
◆ Determine T when x = 10.4 seconds
◆ Determine x when the temperature
of the block of steel rises from 30°C
to 1 500°C.
CHAPTER 4
◆
◆
4 . Making a conclusion
(a) When T = 199.8°C, then
199.8 = 30(1.2)x
199.8 = (1.2)x
30
6.66 = (1.2)x
log 6.66 = x log 1.2
x = log 6.66
log 1.2
= 10.4 seconds
(b) When x = 21.4567 seconds, then
T = 30(1.2)21.4567
≈ 1 500ºC
2 . Planning the strategy
Substitute the value of x into the
formula to find the value of T.
◆ Substitute the value of T into the
formula to find the value of x.
◆
3 . Implementing the strategy
(a) T = 30(1.2)x
= 30(1.2)10.4
= 199.8ºC
Thus, the temperature of the steel after
10.4 seconds is 199.8ºC.
T = 30(1.2)x
(b)
1 500 = 30(1.2)x
1500 = (1.2)x
30
50 = (1.2)x
log 50 = x log 1.2
x = log 50
log 1.2
= 21. 4567
Thus, the time taken for the block
of steel to reach a temperature of
1 500ºC is 21.4567 seconds.
Self Practice 4.13
1. Solve the following equations by giving answers in three decimal places.
(b) 52x − 1 = 79x − 1
(c) 73x − 1 = 50x
(a) 42x − 1 = 7x
2. Solve the following equations by using natural logarithms. Give the answer in three decimal
places.
(c) 5e3x − 4 = 35
(a) ln (5x + 2) = 15
(b) 30e2x + 3 = 145
2x
(f) ln (x + 1)2 = 4
(d) ln (3x – 2) = 4
(e) 41 – e = 5
()
n
3. The price of a house after n years is given by RM260 000 9 . Determine the minimum
8
number of years for the price of the house to exceed RM300 000 for the first time.
120
4.3.5
Indices, Surds and Logarithms
4. A company's savings after n years is RM2 000(1 + 0.07) . Determine the minimum number of
years for their savings to exceed RM4 000.
n
5. After n years, Mr. Chong's money in a bank is RM4 000(1.1)n. Calculate the number of years
for Mr. Chong's money to exceed RM5100 for the first time.
6. The air pressure, in Hg, at a height of 10 km above sea level is given by P = 760e– 0.125h, where
h is the height, in km, and e = 2.718. Determine the height above sea level if the pressure at
that height is 380 mm Hg.
Scan the QR code or visit bit.ly/2osOqUo for the quiz
1. Given log5 3 = 0.683 and log5 7 = 1.209. Without using a calculator or four-figure tables,
calculate log5 1 and log7 75.
( )
2. Given loga 3 = x and loga 5 = y, express loga 453 in terms of x and y.
a
r.
3. Determine the value of log4 8 + logr !w
4. Without using a calculator or four-figure tables, simplify
log12 49 × log64 12
.
log16 7
5. Given log10 x = 2 and log10 y = −1, prove that xy – 100y2 = 9.
6. Given log5 2 = m and log5 7 = p, express log5 4.9 in terms of m and p.
7. Simplify log2 (2x + 1) – 5 log4 x2 + 4 log2 x.
8. Given that log2 xy = 2 + 3 log2 x – log2 y, express y in terms of x.
( )
9. Given log2 b = x and log2 c = y, express log4 8b in terms of x and y.
c
( )
10. The intensity of a sound, in decibel, is calculated by using the formula d = 10 log10 P
P0
where d is the intensity of sound, in decibel, P is the intensity of sound, in Watt and P0 is
the weakest intensity of sound that can be detected by the human ears, in Watt and it is a
constant. In a house, a hot water pump has an intensity of sound of 50 decibels and a wattage
of 10−7 Watts while a dishwasher has a sound intensity of 62 decibels.
(a) Calculate the value of P0.
(b) Determine the wattage, in Watts, for the dishwasher and the hot water pump.
(c) A wattage for sound that exceeds 100 Watts is said to be painful to the human ears. State
the minimum intensity of sound, in decibel, that is considered to be painful to the human
ears.
11. The population growth in a certain country is P = 2 500 000e0.04t where t is the number of
years after year 2020 and e = 2.718.
(a) What is that country's population in 2020?
(b) What is that country's population in 2030?
(c) In which year will that country's population exceed 50 000 000?
4.3.5
121
CHAPTER 4
Intensive Practice 4.3
4.4
Applications of Indices, Surds and Logarithms
Solving problems involving indices, surds and logarithms
Example 35
Mathematics
Application
CHAPTER 4
An entomologist found that a grasshopper infestation towards plants spreads across an area of
A(n) = 1 000 × 20.2n acres, where n is the number of weeks after the initial observation.
(a) Determine the area of infestation at the beginning.
(b) Determine the area of infestation after
(i) 5 weeks,
(ii) 10 weeks.
(c) How much time is needed for the infestation to spread across an area of 8 000 acres?
Solution
1 . Understanding the problem
◆
◆
◆
Given A(n) = 1 000 × 20.2n
n = 0, n = 5, n = 10
A = 8 000 acres
4 . Making a conclusion
(a) When A = 1 000,
1 000 = 1 000 × 20.2n
20.2n = 1
0.2n log 2 = log 1
n = log 1
0.2 × log 2
n = 0 weeks
(b) (i) When A = 2 000,
2 000 = 1 000 × 20.2n
20.2n = 2
0.2n log 2 = log 2
n = log 2
0.2 × log 2
n = 5 weeks
(ii) When A = 4 000,
4 000 = 1 000 × 20.2n
20.2n = 4
0.2n log 2 = log 4
n = log 4
0.2 × log 2
n = 10 weeks
(c) When n = 15,
A = 1 000 × 20.2(15)
= 8 000 acres
122
2 . Planning the strategy
Substitute the value of n into the
given formula.
◆ Substitute the value of A into the
given formula.
◆
3 . Implementing the strategy
(a) A(n) = 1 000 × 20.2n
A(0) = 1 000 × 20.2(0)
= 1 000 × 1
= 1 000 acres
(b) (i) A(n) = 1 000 × 20.2n
A(5) = 1 000 × 20.2(5)
= 1 000 × 21
= 2 000 acres
(ii)A(n) = 1 000 × 20.2n
A(10) = 1 000 × 20.2(10)
= 1 000 × 22
= 4 000 acres
(c) 8 000 = 1 000 × 20.2n
20.2n = 8
20.2n = 23
0.2n = 3
n = 15
Thus, the time taken for the infestation
to spread across an area of 8 000 acres
is 15 weeks.
4.4.1
Indices, Surds and Logarithms
Self Practice 4.14
1. A gardener observes a bug infestation towards plants in his garden. He finds out that the area of
insect infestation towards his plants is A = 1 000 × 20.7n acres, where n is the amount of weeks
after the first week of initial observation. How long will it take for the insects to infest an area
of 5 000 hectares?
Intensive Practice 4.4
Scan the QR code or visit bit.ly/311IqiH for the quiz
1. Mr. Ramasamy keeps RM1 000 in a bank. The amount of money rises by W = 1 000(1.09)t
after t years. Calculate
(a) the amount of money after 5 years,
(b) the time taken, t, in years, for the money to rise from RM1 000 to RM1 200.
2. The remaining radioactive substance of uranium after t years is W(t) = 50 × 2 −0.0002t gram,
where t  0.
(a) Determine the initial mass of the uranium.
(b) Determine the time that is needed for the uranium to weigh 8 grams.
3. The mass, J of a bacteria after time t, in hours is J = 25 × e0.1t gram.
(a) Show that the time taken for the bacterial mass to reach 50 grams is 10 ln 2 hours.
(b) Determine the time taken in two decimal places.
Summary OF Chapter 4
INDICES, SURDS AND LOGARITHMS
Indices
Surds
Laws of Indices
am × an = am + n
am an = am – n
(am)n = amn
4.4.1
• a× a=a
• a × b = ab
a a
• –– = –
b b
• (a + b ) × (a – b )
= a2 – b
Logarithms
• logaxy = logax + logay
x = log x – log y
• loga ––
a
a
y
• logabn = n logab
logcb
• logab = –––––
logca
1
• logab = –––––
logba
123
CHAPTER 4
2. The electric current that flows in an electrical circuit for t seconds after its switch is turned off
is I = 32 × 4−t amp.
(a) Calculate the current flow when the switch is off.
(b) Calculate the current flow after
(i) 1 second,
(ii) 2 seconds.
(c) How long will it take for the current to reach 0.5 amps?
WRITE YOUR JOURNAL
Make a poster that contains all the laws of indices, surds and logarithms according to your
creativity. Each stated law must be accompanied with an example of its usage. Then, hang your
poster in the classroom.
MASTERY PRACTICE
CHAPTER 4
1. Solve the equation 42x − 1 + 42x = 4.
PL1
2. Solve the equation 5n + 1 – 5n + 5n − 1 = 105.
PL2
3. If !w
5 x = !w
3 x + !w
7 , find the value of x in the form of
a
!w
.
b
4. If logx a + logx 1 = t, what is the possible value(s) of t?
a
PL2
PL2
5. The diagram below shows three circles. Circle A has a radius of 2 cm and circle B has a
radius of 1 cm.
A
P
B
Q
PQ is a common tangent and all circles touch one another. Find the radius of the smallest
circle. PL5
6. The temperature of a type of metal decreases from 100°C to T °C according to T = 100(0.9)x
after x second. Calculate PL4
(a) the temperature of the metal after 5 seconds,
(b) the time taken, x, in seconds for the temperature of the metal to decrease from 100°C to
80°C.
()
n
7. After n years, the price of a car that was bought by Raju is RM60 000 7 . Determine the
8
number of years for the price of the car to be below RM20 000 for the first time. PL4
8. Given logx 3 = s and log!wy 9 = t, express log9 x3y in terms of s and/or t.
124
PL4
Indices, Surds and Logarithms
9. Two experiments were carried out to find the relationship between the variables x and y. Both
experiments showed that the relationship between x and y is in accordance to 3(9x) = 27y and
log2 y = 2 + log2 (x − 2). Find the value of x and y that satisfy both experiments. PL5
( )
10. The price of a car drops and can be determined with the equation x log10 1 – 2 = log10 p – log10 q.
y
In this equation, the car with y years of usage and price RMq will drop to RMp after being
used for x years. A car is bought at RM100 000 has 20 years of usage. If the price of the car
drops to RM10 000, find the years of usage for that car. PL5
Exploring
MATHEMATICS
CHAPTER 4
Building an index and surd game with Tarsia software.
1. Download Tarsia software from bit.ly/2SssDGz.
2. Click on “Standard Rhombus Jigsaw” on the following window.
3. Type the question and answer in the relevant spaces. The number of questions that is need
to be completed are shown on the right part of the screen.
Type your question in this space
Example of queston: 42x + 3 = 1 024
Type your answer in this space
Example of answer: x = 1
4. Then, click on the “Output” button on the bottom part of the screen to generate the Jigsaw
Puzzle. Print the Jigsaw Puzzle and cut it according to shape.
5. The Jigsaw Puzzle is ready to be used. Click on the "Solution" button to check the answers.
125
CHAPTER
Progressions
CHAPTER 5
5
What will be learnt?
Arithmetic Progressions
Geometric Progressions
List of
Learning
Standards
bit.ly/2nGImr6
wORDS
KEY W
Sequence
Arithmetic progression
Common difference
Geometric progression
Common ratio
Sum to infinity
Recurring decimal
126
Jujukan
Janjang aritmetik
Beza sepunya
Janjang geometri
Nisbah sepunya
Hasil tambah
ketakterhinggaan
Perpuluhan berulang
5.1.1
Progressions
Did you
Know?
Carl Friedrich Gauss is a mathematician who
is known as the Prince of Mathematics. His
intelligence was proven since he was a child.
Carl Friedrich Gauss corrected his father’s
wage calculation at 3 years old. At the age
of 7, he was able to calculate the sum of 1 to
100 quickly and accurately.
For further information:
CHAPTER 5
bit.ly/2p6owFX
SIGNIFICANCE
OF THIS CH
APTER
The knowledge in solving progression problems
is very important in the field of engineering,
medicine, technology and economy. The
knowledge in progression allows you to determine
the total sums of large numbers with ease.
Bukit Jalil National Stadium is the largest
stadium in Malaysia. It has indoor shaded
seats for the spectators. How do we know
the total number of seats without counting
it one by one? How did the number of
seats increase from the innermost row
to the outermost row? Can you form an
equation to calculate the total number of
seats in the stadium?
Scan this QR code to
watch video of Bukit Jalil
National Stadium.
bit.ly/2Vijima
5.1.1
127
5.1
Arithmetic Progressions
Identifying arithmetic progressions
Mr. Lee built stairs in his garden. He used eight bricks on the first level. For each subsequent
level he increased another 8 bricks. The total number of bricks used on each level can be
written in a progression of 8, 16, 24, … If Mr. Lee wanted to build 18 steps, how many bricks
would be needed?
8, 16, 24, … is a finite sequence that follows a specific pattern. Sequence such as 3, -3, 3,
-3, … is an infinite sequence. Each number in a sequence is known as terms, such that the first
term is written as T1, second term T2 and so on until Tn, which is the nth term.
INQUIRY 1
In groups
Aim: To understand arithmetic progressions
CHAPTER 5
Instruction:
1. Observe the following polygons in which the number of sides of the consecutive
polygons increase by one.
(a)
(b)
(c)
(d)
(e)
(f)
2. Divide each polygon into triangular-shaped as shown in (b) and (c).
3. In the table, fill in the sum of interior angles for each of the given polygon.
Polygon arrangement, n
n=1
Sum of interior angles
180°
n=2
n=3
n=4
n=5
n=6
4. How do you obtain consecutive terms for the sum of interior angles in the polygons?
5. Explain the relationship between any two consecutive terms and state the finite value
that relates the two terms.
6. Without plotting a graph, find the sum of interior angles for the tenth polygon
arrangement.
From the results in Inquiry 1, it is given that the difference between any two consecutive
terms of a sequence is a fixed constant. The constant is known as common difference and is
represented by d. Thus:
d = T 2 – T 1 = T 3 – T 2 = … = Tn – T n – 1
128
d ≠ T1 – T2 ≠ T2 – T3 ≠
… ≠ Tn – 1 – Tn
5.1.1
Progressions
Sequence that has common difference, d is known as arithmetic progression.
Arithmetic progression is a sequence of numbers such that each term is obtained by
adding a constant to the term before it.
Example 1
Determine whether each of the following sequence is an arithmetic progression.
Give your justification.
2
10
(b) , 2, , 5, …
(a) 358, 350, 342, …
3
3
Solution
2 4
(a) d1 = 350 – 358 = –8
(b) d1 = 2 – =
3 3
d2 = 342 – 350 = –8
10
4
= –2=
d
This sequence is an
2
3
3
arithmetic progression
10 5
=5– =
d
because d1 = d2 = –8.
3
3 3
CHAPTER 5
This sequence is not an arithmetic
progression because
d1 = d2 ≠ d3.
Example 2
An auditorium has 15 chairs in the first row, 19 chairs in the
second row, 23 chairs in the third row and so on. Determine
whether the arrangement of chairs in each row follows an
arithmetic progression. Give your justification.
Solution
Sequence: 15, 19, 23, …
d1 = 19 – 15 = 4
d2 = 23 – 19 = 4
Since, the difference of the progression is a fixed constant, which is 4. Thus, the arrangement
of chairs in each row in the auditorium follows an arithmetic progression.
Self Practice 5.1
1. Find the common difference for each of the following arithmetic progression and state the
method to obtain the arithmetic progression.
3, 5!w
3, 8!w
3, …
(a) –35, –21, –7, …
(b) 2!w
(c) p + q, 2p, 3p – q, …
(d) loga 2, loga 24, loga 27, …
5.1.1
129
2. Determine whether each of the following sequence is an arithmetic progression and give your
justification.
1 1 1 1
(a) 9, 13, 17, 21, …
(b) , , , , …
2 4 6 8
(d) 5 – x, 5, 5 + x, …
(c) 0.1, 0.01, 0.001, …
3. Complete the network diagram below, given that the relationship of each of the following
network is a consecutive term in an arithmetic progression.
(a)(b)(c)
10
5
7x
–p
12p
8
CHAPTER 5
12x
2p
5x
4. Azrul and Jonathan are placing national flags along the school corridor from the canteen to
the staffroom. The distance between the first and second flag is 5m. The third flag is placed
10 m from the first flag and this is the pattern of arrangement for the rest of the flags until the
end. Determine whether the arrangement of these flags follows arithmetic progression. Give
justification for your answer.
Deriving the formula of the nth term, Tn of arithmetic progression
INQUIRY 2
In groups
Aim: To derive the formula of the nth term, Tn of arithmetic progression
Instruction:
1. Consider an arithmetic progression 2, 5, 8, 11, 14, … Use the pattern of this sequence to
complete the table.
2. Assume the first term of an arithmetic progression is a and the common difference is d.
3. Complete the table below.
Term
T1
T2
T3

Tn
Value of
term
a
a +d
a + d +d

Method to obtain the value of term
Formula
(deduction method)
Does not have d
T1 = a + 0d
Add d at T1 term
T2 = a + 1d
Add d at T2 term
T3 = a + 2d


4. In your group, answer the following questions.
(a) Express T20, in terms of a and d.
(b) State the relationship between the term Tn and its common difference.
(c) Write a general formula for Tn.
130
5.1.1
5.1.2
Progressions
From the results of Inquiry 2, the nth term of an arithmetic progression can be written as:
Tn = a + (n – 1)d
Such that a is the first term, d is the common difference and n is the number of terms.
Quick Count
(a) Find the 15 term for the arithmetic progression – 4, 2, 8, …
(b) Find the 24th term for the arithmetic progression – 6, 5, 16, …
th
Solution
(a) First term, a = – 4
Common difference, d = 2 – (– 4) = 6
15th term, T15 = – 4 + (15 – 1)6
= 80
(b) First term, a = – 6
Common difference, d = 5 – (– 6) = 11
24th term, T24 = – 6 + (24 – 1)11
= 247
Example 4
Based on Example 3, we can
use scientific calculator to
obtain the 15th term.
1. Press –4 + (
ALPHA )
– 1
) ( 6 ) CALC
The screen displays:
–4 + (x – 1)(6)
x=
2. Press 15 =
The screen displays:
–4 + (x – 1)(6)
80
3. Press = to enter the
value of other terms.
Given that an arithmetic progression with the first term is -6, the
common difference is 11 and the nth term is 126, find the value
of n.
Solution
a = – 6, d = 11, Tn = 126
Tn = a + (n – 1)d
126 = – 6 + (n – 1)(11)
126 = 11n – 17
n = 13
Example 5
In a book fair, Siti wants to arrange books at the front section.
She arranges the books by stacking them with the first book at
2 cm thickness at the bottom. However, the subsequent books
have the same thickness, which is 1.5 cm. Find
(a) the total thickness of the books if Siti arranged 16 books,
(b) the number of books if the height of the books is 30.5 cm.
5.1.2
131
CHAPTER 5
Example 3
Solution
(a) The sequence of the thickness of books: 2, 3.5, 5, 6.5, …
a = 2, d = 1.5
Total thickness of books at the 16th book. = 2 + (16 – 1)(1.5)
= 24.5 cm
Thus, the total thickness of 16 books is 24.5 cm.
(b) Tn = 30.5
30.5 = 2 + (n – 1)(1.5)
n – 1 = 19
n = 20
Thus, there are 20 books.
Self Practice 5.2
1. Find the way to the “FINISH” box by choosing the correct answer.
START
3x – 1, 4x, 6x – 2
are three consecutive
terms. Find T6.
CHAPTER 5
Find the 9th term
in the sequence
9, 5, 1, …
–23
25
FINISH
24
–18
3
Given that a = 10
and d = – 4, find
T7.
40
Given that T2 = 3 and
T17 = 54, find the first
term, a.
Given that Tn = 8 – 3n.
Find the common
difference, d.
Given that
Tn = 8 – 5n, find T4.
–62
8x – 4
T5 = 20 and
T4 = 2T2, find T10.
–4
– 0.4
Given that –17, –14,
–11, … 55, find the
number of terms.
2. Encik Muiz starts to work in a company at a certain month. His first annual income is
RM36 000 and the increment for the following year is RM1 000. Calculate
(a) the number of years in which Encik Muiz needs to work in order for him to receive twice
his first annual income,
(b) the yearly increment if his salary during the 6th year is RM43 500.
132
5.1.2
Progressions
Deriving the formula of sum of the first n terms, Sn , of the
arithmetic progression
INQUIRY 3
In groups
Aim: To derive the formula of sum of the first n terms, Sn, of the arithmetic progression
Instruction:
1. Observe the table below.
S2
Number of grids based on
Formula of a rectangle
the number of terms
by deduction method
Area of a rectangle
Diagram II
Diagram I
T1 = a
= (T1 + T2)2
T2 = a + (2 – 1)d
d
T1
=
[a + a + (2 – 1)d]2
T2
=a+d
T2
a T1
2[2a + (2 – 1)d]
T2
S2 =
T
1 unit 1 unit
2
1
Diagram III
T1 = a
Diagram IV
T2 = a + (2 – 1)d
T2
T3 = a + (3 – 1)d
T3
S3
T1
S4

Sn
T2
T3
T1

T2
Area of a rectangle
= (T1 + T3)3
(T + T )3
S3 = 1 2 3
3
S2 = [a + a + (3 – 1)d]
2
T1
T3

2. Diagram I shows two grids with 1 unit width arranged side by side.
• The height of blue grid is a unit which is represented by the first term T1.
• The height of red grid is d unit longer than that of the blue grid that represent the
second term, T2 = a + d or T2 = a + (2 – 1)d.
3. In Diagram II, the red grid is placed on top of the blue grid so that the total height is
T1 + T2 = a + a + (2 – 1)d unit. The blue grid is placed on top of the red grid so that the
height becomes T1 + T2 = a + a + (2 – 1)d unit.
4. Observe that both the blue and red grids become a rectangle. The sum of the blue grid
2[2a + (2 – 1)d]
and red grid, S2, is half of the area formed. The sum can be written as
.
2
5. Repeat step 1 to 3 to find S4 and Sn.
6. Derive the formula of sum of the first n terms, Sn.
From the results of Inquiry 3, we found that the sum of the first n terms, Sn, of the arithmetic
progression is derived by using the area of rectangle formed from the terms of the arithmetic
progression.
Therefore, the sum of the first n terms, Sn, can be written as:
Sn =
n
[2a + (n – 1)d]
2
Such that a is the first term, n is the number of terms and d is the common difference.
5.1.3
133
CHAPTER 5
Sum of
terms
Since Tn = a + (n – 1)d is also a last term, l, hence the sum of the nth term, Sn can be expressed
as follow:
Sn =
n
[a + Tn]
2
or
Sn =
n
[a + l]
2
The nth term of an arithmetic progression can be obtained by the formula of sum of the first
n terms, Sn. For example, to find the value of 10th term in an arithmetic progression, the sum of
the first ten terms minus the sum of the first nine terms, which is T10 = S10 – S9. In general:
Tn = Sn – Sn – 1
Example 6
Given that an arithmetic progression 4, 7, 10, …, find
(b) the sum of the first nth terms.
(a) the sum of the first 35 terms,
CHAPTER 5
Solution
(a) First term, a = 4
Common difference, d = 7 – 4 = 3
S35 = T1 + T2 + T3 + … + T35
35
S35 =
[2(4) + (35 – 1)(3)]
2
= 1 925
Example 7
n
[2(4) + (n – 1)(3)]
2
n
= [5 + 3n]
2
(b) Sn =
The sum of the first ten terms of an arithmetic progression is
230 and the sum of the subsequent ten terms is 630. Find the
first term, a and the common difference, d for this arithmetic
progression.
Mind Challenge
In Example 7, why is
S20 = 230 + 630?
Explain your answer.
Solution
10
[2a + (10 – 1)d]
2
230 = 5(2a + 9d)
46 = 2a + 9d … 1
20
[2a + (20 – 1)d]
S20 =
2
230 + 630 = 10(2a + 19d)
860 = 10(2a + 19d)
86 = 2a + 19d … 2
2 – 1 : 40 = 10d
d=4
S10 =
134
BRAINSTORMING
Discuss with your friends
and prove that:
(a) S8 – S5 = T6 + T7 + T8.
(b) Sn – Sn – 1 = Tn.
5.1.3
Progressions
Substitute d = 4 into
46 = 2a + 9(4)
2a = 10
a=5
1
Smart TIPS
,
Arithmetic progression is
written in the form of
T1, T2, T3, … whereas
arithmetic series is written in
the form of T1 + T2 + T3 + …
Thus, the first term, a is 5 and the common difference, d is 4.
Example 8
Solution
(a) The sequence of the number of hexagonal holes: 2, 5, 8, …
This sequence is an arithmetic progression.
Common difference, a = 2
First term, d = 5 – 2 = 3
Total number of hexagonal holes on the 12th day,
12
S12 =
[2(2) + (12 – 1)(3)]
2
= 222
(b) Total number of days, Sn = T1 + T2 + T3 + … Tn
Sn > 1 000
n
[2a + (n – 1)d] > 1 000
2
n
[2(2) + (n – 1)(3)] > 1 000
2
n[1 + 3n] > 2 000
3n2 + n > 2 000
3n2 + n – 2 000 > 0
+
–
–25.99
25.65
–25.99 < n < 25.65
n < –25.99
+
x
5.1.3
The beehive is made of a
combination of hexagonal
prismatic cells so that there
are no space formed in
between the hexagonal
shapes. Thus, bees do not
need to use a lot of wax to
make their hives. The surface
area of the hexagonal shapes
is the largest compared to
other shapes.
Scan the QR code to check
out the reason beehive is
hexagonal.
bit.ly/2AYE4hM
FLASHBACK
If 3n2 + n – 2 000 = 0, then
–1 ± !w
12 – 4(3)(–2 000)
,
2(3)
and n = 25.65 or
n = –25.99
n=
n > 25.65
153.92
155.92
or n < –
6
6
n > 25.65
< –25.99 (Ignore)
Thus, the minimum number of days to make more than
1 000 hexagonal holes is 26 days.
n>
MATHEMATICS
POCKET
Mind Challenge
In Example 8, why is the
value –25.99 not taken into
account?
135
CHAPTER 5
A swarm of bees started to make a new beehive. 2 hexagonal
holes were made on the first day. 5 hexagonal holes were made
on the second day and 8 hexagonal holes were made on the third
day and followed on until the beehive is ready.
Calculate
(a) the number of hexagonal holes on the 12th day,
(b) the minimum number of days if there are more than 1 000
hexagonal holes made.
Self Practice 5.3
1. Find the sum of the following arithmetic progression.
3 6 9
(a) –20, –15, –10, …, 100 (b) , , , … till the first 23 terms.
5 5 5
2. Complete the crossword puzzle.
CHAPTER 5
Horizontal:
(a) Find the sum of the arithmetic series 38 + 34 + 30 + …
till the first 18 terms.
(b) Find the sum of the first 100 terms of an arithmetic
progression such that the first term is –10 and the
common difference is 6.
(a)
(c) Find the first term of the arithmetic progression in
which the sum of the first 42 terms is 5 838 and the
last term is –22.
(e)
(c)
(d)
(b)
Vertical:
(c) Calculate S140 of an arithmetic progression that has 140 terms in which the first term and
the last term are 2 and 449 respectively.
(d) Calculate the value of n of an arithmetic progression as such that the first term is –15, the
common difference is –3 and the sum of the first n terms is –1 023.
(e) Calculate the sum of 200 terms after the first 50 terms of an arithmetic progression in
n
which the sum of the first n terms is Sn = [n + 1].
2
3. The diagram on the right shows the pattern drawn on a
Cartesian plane. The final line on the plan is parallel to the
y-axis and passes through x = –10. Find the sum of the length
of the overall pattern.
y
4
3
2
1
–4 –3 –2 –1 0
–1
–2
–3
–4
1 2 3 4
x
4. The diagram on the right shows a fence
that is made of wood. The fence is painted
with alternate blue and grey as shown in the
diagram. The number of wood pieces painted
with the same colour increases with the rate
as shown in the diagram. If there are only
200 wood pieces,
(a) calculate the number of wood pieces that can be painted with the same colour. Hence,
find the number of remaining wood pieces, if any.
(b) state the colour of the last wood piece and then, calculate the number of the wood pieces
used for that particular colour.
136
5.1.3
Progressions
Solving problems involving arithmetic progressions
Example 9
MATHEMATICS APPLICATION
Encik Suhaimi is a chicken breeder that has 1 500 chickens.
He plans to sell 200 chickens each day. He gives food to all the
chickens with the expenses of RM0.50 per chicken in a day.
Calculate the total expenses of the food spent by Mr Suhaimi if
he starts with 1 500 chickens until he is left with 300 chickens.
Solution
◆
Find the total expenses of chicken food
until there are 300 chickens left.
2 . Planning a strategy
Form an arithmetic progression
series with a first term, a and common
difference d, until the last term, 300.
◆ Determine the number of days for
Encik Suhaimi to be left with 300
chickens by using the formula
Tn = a + (n – 1)d.
◆ Determine the total expenses of food
when there are 300 chickens left by
using the formula
n
Sn = [2a + (n – 1)d].
2
◆
3 . Implementing the strategy
Arithmetic progression:
1 500, 1 300, 1 100, …, 300
First term = 1 500
Common difference = –200
Tn = a + (n – 1)d
300 = 1 500 + (n – 1)(–200)
300 = 1 700 – 200n
200n = 1 400
n =7
On the 7th day, the number of chickens left
is 300.
7
S7 = [2(1 500) + (7 – 1)(–200)]
2
= 6 300
The total expenses of food is
= 6 300 × RM0.50
= RM3 150
4 . Making a conclusion
n = 7, T7 = 1 500 + (7 – 1)(–200)
= 300
5.1.4
137
CHAPTER 5
1 . Understanding the problem
Self Practice 5.4
1. Mr. Tong ordered 1 000 Form 4 Mathematics textbook to be sold at his shop. He estimated
that 10 books would be sold on the first day, 14 books on the second day and 18 books on the
third day and the following day with the same increment.
(a) Calculate the number of days for Mr. Tong to sell all the books.
(b) Calculate the increment of the books to sell each day in order for all the books to be sold
in 10 days.
2. A wire with length of 240 cm is cut into 15 pieces such that the length of each piece follows
the arithmetic progression. The longest piece of the wire is 30 cm.
(a) Calculate the length of the shortest wire.
(b) Find the difference between two consecutive wires.
CHAPTER 5
Intensive Practice 5.1
Scan the QR code or visit bit.ly/2p3S7zY for the quiz
1. Determine whether the following sequence is an arithmetic progression and give justification
for your answer.
(a) –32, –17, –2, 13
(b) 8.2, 5.7, 3.2, 1.7, – 0.8
2. For each of the following arithmetic progression, find the nth term as stated in the bracket.
1
1
(a) –12, –9, – 6, … [9th term]
, – , –1, … [15th term]
(b)
3
3
3. Determine the number of terms for each of the following arithmetic progression.
(a) – 0.12, 0.07, 0.26, …, 1.97
(b) x, 3x + y, 5x + 2y, …, 27x + 13y
4. Find the sum of arithmetic progression –23, –17, –11, … which
(a) has 17 terms,
(b) has 2n terms, in terms of n
(c) the last term is 121.
5. Given that Sn = 2n2 – 5n, find
(a) the first term,
(b) the 9th term,
(c) the sum of the 4th term to the 8th term.
6. The 2nd term of an arithmetic progression is
(a) the common difference,
1
and the sum of first 14th terms is –70. Find
2
(b) the last term.
7. Yui Ming received offers to work in two companies with the following income.
Company A: Monthly income is RM3 500 and the increment is RM20 per month.
Company B: Annual income is RM46 000 and the increment is RM1 000 per year.
Yui Ming plans to work for 3 years. Which company is more suitable for her in order to
receive maximum income in 3 years? Show your calculation and calculate the difference of
the total income between the two companies.
138
5.1.4
Progressions
5.2
Geometric Progressions
Identifying geometric progressions
There is a famous legend regarding the invention of chess related
to series. According to the legend, a king from India wanted to
meet the chess inventor to give recognition because this invention
was interesting. The chess inventor only requested wheat to be
given to him according to the following calculation:
1 grain of wheat in the first grid, 2 grains of wheat in the second
grid, 4 grains of wheat in the third grid and so on until the last
grid.
INQUIRY 4
In groups
CHAPTER 5
When the chess board was filled, the total amount of grains given to the chess inventor was
1.84 × 1019, which is equivalent to 1.2 metric tons. The calculation of the amount of grains can
be obtained using geometric progression.
21st Century Learning
Aim: To identify geometric progression
Instruction:
1. Read the situation below carefully.
There are various bacteria in our surroundings. Bacteria can be found on unhygienic
food, human and animal intestines. Bacteria can reproduce very fast and lead to
diseases such as diarrhoea. The rate at which a bacterium reproduces is double at
the same rate, which is every 20 minutes: a bacterium will become two, two bacteria
will become four and so on. If a human intestine has 2 million bacteria, he/she will
be infected with diarrhoea.
2. Let’s say if a type of food has only one bacterium. If you consume the food, estimate the
time taken for you to be infected with diarrhoea.
3. The table below shows the number of bacteria reproduced. One box represents the
reproduction of the bacteria in 20 minutes. Complete the table below until the number
of bacteria leads to diarrhoea.
1
5.2.1
2
4 = 22
8 = 23
139
4. How long does it take for you to be infected with diarrhoea?
5. Determine the method to get the number of bacteria every 20 minutes from the
previous 20 minutes. Is the value a constant?
6. Use the GeoGebra software and draw a graph to represent the number of bacteria
increasing with time.
7. Discuss with your group members about the results obtained and record the results
obtained on a piece of paper.
8. Each group moves to another group to compare the results obtained.
CHAPTER 5
Geometric progression is a sequence of numbers where each
term is obtained by multiplying a constant with the previous
term.
If T1, T2, T3, …, Tn, is the first n terms of a geometric
progression. The ratio of the two consecutive terms is called the
common ratio, r.
r=
T2 T3
T
= =…= n
T1 T2
Tn – 1
r≠
T1 T2
T
≠ ≠ … ≠ n–1
T2 T3
Tn
Example 10
Determine whether the progression below is a geometric
progression. Give your justification.
(a) 5, 15, 45, 135, …
(b) 0.1, 0.2, 0.3, …
Solution
MATHEMATICS
POCKET
The graph for geometric
progression is quite similar to
exponential function graph.
Geometric progression graph
is discrete while exponential
function graph is parallel.
Graph of geometric
progression
Number of boxes
From the results of Inquiry 4, it is found that the ratio between any
two consecutive terms is a fixed number. So, the progression is
called geometric progression.
80
60
40
20
0 1 2 3 4
Number of arrangement
Graph of exponential
function
y
10
8
6
4
2
–6 –4 –2 0
2 4
x
15
45
135
= 3, r2 =
= 3, r3 =
=3
5
15
45
This progression is a geometric progression because the
common ratio, r is the same.
0.2
0.3 3
(b) r =
= 2, r =
=
0.1
0.2 2
This progression is not a geometric progression because the
common ratio, r is different.
(a) r1 =
140
5.2.1
Progressions
Self Practice 5.5
1. Determine whether the following sequence is a geometric progression. Justify your answer.
40
(a) 120, 40, , …
3
(b) 0.03, 0.003, 0.0003, …
(c) x + 1, 2x, 5x + 12, 12x, …
2. Complete the network diagram below, given that the relationship of each of the following
network is a consecutive term in a geometric progression.
(b)
(a)
1
1
1
1
3
12
24
8
1
12
3. Given that x – 2, x + 1, 4x + 4 are three consecutive terms in a geometric progression, state
the positive value of x. Hence, list the first three terms and state the common ratio.
Deriving the formula of the n th term, Tn , of geometric progressions
INQUIRY 5
In groups
Aim: To derive the formula of the nth term, Tn, of geometric progressions
Instruction:
1. Consider the geometric progression 2, 6, 18, 54, … with the first term, a and the
common ratio, r.
2. Discuss with your group members and complete the table below.
Term
Value of term
T1
2
6
T2
T3
Method to obtain the value of term
2(3)
1–1
2(3)
2–1
Formulae
= 2(3)
0
a
= 2(3)
1
ar = ar2 – 1
18
T4
54
T5




Tn
3. Derive a formula of the nth term of geometric progressions.
5.2.1
5.2.2
141
CHAPTER 5
1
2
2
From the results of Inquiry 5, we can notice that the value of each term in this geometric
progression can be obtained by using the formula below.
Tn = arn – 1
Such that a is the first term, r is the common ratio and n is the number of terms.
Example 11
(a) Find the common ratio and the 5th term of the geometric progression 4, –20, 100, –500, …
2 2 2
(b) Find the common ratio and the 7th term of the geometric progression 2, , , , …
3 9 27
Solution
(a) First term, a = 4
(b) First term, a = 2
2
1
–20
Common
ratio, r = ÷ 2 =
= –5
Common ratio, r =
3
3
4
T5 = 4(–5)5 – 1
1 7–1
T7 = 2
3
= 2 500
2
=
729
CHAPTER 5
()
Example 12
Find the number of terms of the geometric progression –
Solution
First term, a = –
Tn = ar
n–1
25 5 1
1
, , – , …,
.
3 3 3
9 375
( )
25
5
25
1
, common ratio r = ÷ –
=–
3
3
3
5
( )( )
( )
( ) ( )
1
25
1 n–1
= –
–
9 375
3
5
1
1 n–1
= –
–
78 125
5
1 7
1 n–1
–
= –
5
5
7=n–1
n=8
Thus, the number of terms is n = 8.
Example 13
An open stadium has 20 chairs in the first row. The number of chairs in the next row is one
and a half times the number of chairs of the previous row.
(a) Calculate the maximum number of chairs in the 10th row.
(b) Which row has at least 505 chairs?
142
5.2.2
Progressions
Solution
20(1.5)n – 1  505
505
(1.5)n – 1 
20
505
(n – 1) log 1.5  log
20
505
log
20
n–1 
log 1.5
n  7.96 + 1
n  8.96
(b)
(a) First term, a = 20
Common ratio, r = 1.5
Sequence in geometric progression:
20, 30, 45, …
T10 = 20(1.5)9
= 768.9
Thus, the maximum number of chairs
in the 10th row is 768.
Thus, the 9th row has at least 505 chairs.
Self Practice 5.6
START
T1 = 12 and
T3 = 27, find T5.
If T1 = 4 and
T3 = T2 + 24,
find the positive
values of r.
3
Given that a = 50 and
12
T5 = 20 , find r.
25
3
5
8
Given that T2 =
9
9
and T6 = . Find T5.
2
Given that –3, 6,
–12, …, −192,
find the number
of terms.
CHAPTER 5
1. Find the way to the “FINISH” box by choosing the correct answer.
FINISH
0.01
Given that
0.12, 0.0012,
0.000012. Find r.
12
1
Given that r = and
2
1
T3 = . Find T10.
6
Find T5 for the
sequence x + 1,
x + 3, x + 8, …
2. The diagram on the right shows a ball bouncing on the floor.
The maximum height of the bounced of the ball is 3 m and the
height of each bounce is 95% of the previous bounce. From
which bounce onwards will the height be less than 1 m?
5.2.2
143
Deriving the formula of sum of the first n terms, Sn , of geometric
progressions
Consider a geometric progression with the following terms:
a, ar, ar2, ar3, …, arn – 2, arn – 1
CHAPTER 5
Let’s say the sum of the first n terms is Sn.
Hence, Sn = a + ar + ar2 + ar3 + … + arn – 2 + arn – 1 … 1
1 × r: rS = ar + ar2 + ar3 + ar4 + … + arn – 1 + arn … 2
n
A 1 – 2 : – rS = a + ar + ar2 + ar3 + … + arn – 2 + arn – 1
n
– rSn = ar + ar2 + ar3 + ar4 + … + arn – 1 + arn
All the terms in between a and
Sn – rSn = a – arn
arn is eliminated.
Sn(1 – r) = a(1 – rn)
a(1 – rn)
Commonly used
Sn =
,r≠1
1–r
when |r| < 1
B If 2 – 1 : rS – S = arn – a
n
n
Sn(r – 1) = a(rn – 1)
Commonly used
a(rn – 1)
Sn =
,r≠1
when |r| > 1
r–1
In a geometric progression, the nth term is also calculated by
deducting the sum of (n – 1)th term from the sum of nth term. For
example, given that the geometric progression is 1, –3, 9, –27, …
the 5th term can be calculated by deducting the sum of first 4 terms
from the sum of first 5 terms, which is T5 = S5 – S4. Therefore, the
formula to find Tn by using the sum of terms can be written as:
Tn = Sn – Sn – 1
Smart TIPS
• |r| < 1 can be written as
–1 < r < 1.
• |r| > 1 can be written as
r < –1 and r > 1.
MATHEMATICS
POCKET
1, 2, 4, … is a geometric
progression whereas
1 + 2 + 4 + … is a
geometric series.
Example 14
Given that a geometric series is 1 + 5 + 25 + 125 + 625 + …
(a) Find the sum of first 10 terms.
(b) Find the value of n where Sn = 3 906.
Solution
(a) First term, a = 1
Common ratio, r = 5
a(rn – 1)
Use this formula
Sn =
because|r| > 1
r–1
10
1(5 – 1)
S10 =
5–1
= 2 441 406
144
(b)
Sn = 3 906
1(5n – 1)
= 3 906
5–1
5n – 1 = 15 624
5n = 15 625
n log 5 = log 15 625
log 15 625
n =
log 5
= 6
5.2.3
Progressions
Example 15
A health product company has planned a
marketing strategy. Each member has to
promote the company products by getting
two downlines.
(a) Show that the number of members
in each level follows a geometric
progression.
(b) If there are 9 levels in the marketing
strategy, find the total number of
members involved in promoting the
product.
Level 1
Level 2
Level 3
Level 4
(a) The number of members in each level can be written as 1, 2, 4, 8, …
2 4
r= = =2
1 2
Since r = 2, thus, the number of members in each level follows a geometric progression.
(b) When n = 9, S9 = 1 + 2 + 4 + 8 + … + T9
a(rn – 1)
Use Sn =
r–1
1(29 – 1)
S9 =
2–1
= 511
The total number of members involved in promoting the
product is 511 members.
BRAINSTORMING
By using
n
Sn = a(r – 1) and
r–1
a(rn – 1 – 1)
Sn – 1 =
, prove that
r–1
Tn = arn – 1.
Self Practice 5.7
1. Find the sum of each of the following.
(a) 0.02, 0.04, 0.08, …, T12
(b) p, p3, p5, …, p21, in terms of p
1 3 9
(c) , , , … till the first 15 terms
2 2 2
2. Find the number of terms if the sum of geometric progression 3 500, 700, 140, … is 4 368.
3. A squared paper is cut into 4 equal-sized squares. Each portion is then cut again into 4 smaller
equal-sized portion. This process is repeated on each of the small portion square.
(a) Show that the number of squares cut forms a geometric progression.
(b) Find the total squares obtained if the process is repeated for 6 times.
5.2.3
145
CHAPTER 5
Solution
Determining the sum to infinity of geometric progressions
INQUIRY 6
In groups
Aim: To determine the sum to infinity of geometric progressions
Instruction:
1. Consider the geometric progression 64, 32, 16, …
2. Complete the table on the right for the value of rn and Sn.
CHAPTER 5
3. Discuss with your group members about your observation
on these two values when n increases.
a(1 – rn)
4. Derive a conclusion for Sn =
when n increases to
1–r
infinity. Then express S∞ in terms of a and r.
n
rn
Sn
1
2
3
4
5
10
5. One of the group members will present their findings in
front of the class and the members from the other groups
will ask questions.
100
6. Other groups will take turns to do their presentation.
200
20
From the results of Inquiry 6, when the value of n increases and get closer to infinity (n → ∞),
the value or rn will decrease and get closer to zero (rn → 0) whereas the value of Sn will get
a
a
closer to
S →
. Hence, the sum to infinity of geometric progressions is
1–r n
1–r
(
)
S∞ =
Example 16
a
, where |r| < 1
1–r
Find the sum to infinity of geometric progressions 45, 9, 1.8, …
Solution
a = 45, r =
S∞ = 45
1
1–
5
1
= 56
4
146
9 1
=
45 5
QR
Proof of Pythagoras
theorem using sum to
infinity of geometric
progressions.
bit.ly/2nA0ra4
5.2.4
Progressions
Example 17
Solution
S∞ = 31
1
2
a
63
=
1–r 2
63
(1 – r) … 1
a=
2
a + ar = 28
… 2
a(1 + r) = 28
28
a(1
+
r)
2 ÷ 1 , =
63
a
(1 – r)
2
8
(1 + r)(1 – r) =
9
8
2
1–r =
9
1
r2 =
9
1
1
or r = –
r =
3
3
Example 18
Express the repeating decimals 0.56363… in the form of sum to
infinity of geometric progressions. Hence, express the number in
the simplest form of fraction.
Mind Challenge
1+ 1 + 1 + 1 +…=2
2 4 8
Look at the diagram below
and make a conclusion.
1
1
16
1
8
1
4
1
2
Use the similar diagram and
prove that
2 + 1 + 1 + 1 + … = 4.
2 4
CHAPTER 5
1
The sum to infinity of geometric progressions is 31 and the
2
sum of first two terms is 28. Find the common ratio.
MATHEMATICS
POCKET
The repeating decimals
such as 0.56363… can be

written as 0.563.
Solution
0.56363… = 0.5 + 0.063 + 0.00063 + 0.0000063 + …
= 0.5 + (0.063 + 0.00063 + 0.0000063 + …)
= 0.5 + S∞
1
0.063
= +
2 1 – 0.01
1
7
= +
2 110
31
=
55
5.2.4
Smart TIPS
0.063 + 0.00063 + 0.0000063
+ … is a geometric series
with a = 0.063 and r = 0.01.
147
Self Practice 5.8
1. Complete the following crossword puzzle.
Horizontal:
(a) Find the sum to infinity of geometric progressions
2
1 500, 500, 166 , …
3
(b) Wilson loaned RM15 000 to buy a motorcycle. Every year,
he managed to reduce 50% of his total loan. Find the
maximum amount paid by Wilson.
(a)
(c)
(d)
(b)
Vertical:
(c) Given that the sum to infinity is 4 480 and the common ratio
1
is , find the first term of this geometric progressions.
2
h
(d) 4.818181… can be written in the form of , find the value of h.
11
CHAPTER 5
Solving problems involving geometric progressions
Example 19
A telecommunication company managed to sell 0.5 million smartphones in the year 2015.
Every year, the sales of the smartphone increases by 4%.
(a) Find the total number of smartphones sold from the year 2015 to the year 2020.
(b) If 33% of the smartphones sold from the year 2017 to the year 2020 is 5-inched phones and
14% are 6-inched phones, calculate the total number of 5-inched and 6-inched phones.
Solution
(a) Geometric progression (in million): 0.5, 0.5(1.04), 0.5(1.04)2, …
a = 0.5 million, r = 1.04
0.5(1.046 – 1)
S6 =
1.04 – 1
= 3.316 million
(b) The total number of smartphones from year 2017 to year 2020.
0.5(1.046 – 1) 0.5(1.042 – 1)
S6 – S2 =
–
1.04 – 1
1.04 – 1
= 3.316 million – 1.02 million
= 2.296 million
The number of 5-inched smartphones:
33
× 2.296 million = 0.758 million
100
The number of 6-inched smartphones:
14
× 2.296 million = 0.321 million
100
Total number of smartphones = 0.758 million + 0.321 million
= 1.079 million
Thus, the total number of 5-inched and 6-inched smartphones sold are 1.079 million.
148
5.2.4
5.2.5
Progressions
Self Practice 5.9
1. A wire is cut into a few pieces, in which 10x cm, (4x + 20) cm and (3x – 10) cm are the three
consecutive pieces of a geometric progression.
(a) Find the longest piece if 10x is the 2nd longest term.
(b) If the wire is cut into sum to infinity pieces, find the maximum length of the wire, in m.
2. The diagram on the right shows the pattern of a spider web. The perimeter of each semicircle
follows geometric progression with the smallest radius r cm and
each subsequent radius increases 40%.
(a) Form the first three terms for the perimeter of the semicircle
in terms of r.
(b) Find the total length of semicircle, in m, if the spiderweb
has 15 semicircles and each radius is 2 cm.
Scan QR code or visit bit.ly/2VrxSZ5 for the quiz
1. Calculate the number of terms and the sum for each of the following geometric progression.
(a) –1, 3, –9, …, 2 187
(b) log x–1, log x–2, log x–4, …, log x–64
3 3
3
(c) 0.54, 0.0054, 0.000054, … 5.4 × 10–17
(d) 3, , , …
2 4
64
2. Given that the geometric progression 4.5, –9, 18, … Find the number of terms for this
geometric progression in order for the sum to be 769.5.
3. Three consecutive terms of a geometric progression is x, 2x + 3 and 10x – 3. Find
(a) all the possible values of x,
(b) the 6th term if x < 0.
4. The diagram shows a few triangles. Given that the area
of the triangles follows geometric progression such that
the area of the third triangle is 36 cm2 and the sum of
area of the third and fourth triangle is 54 cm2. Find
(a) the common ratio and the area of the first triangle,
(b) the sum of the third triangle to the tenth triangle.
...
5. The diagram shows a few circles with the same centre. The perimeter
of each circle follows the geometric progression. Given that the
nth perimeter is Tn = 38 – n cm, find
(a) the common ratio,
(b) the sum of three consecutive perimeter after the second largest
perimeter.
6. There are three children with their mass arranged according to the descending order of a
geometric progression. The sum of mass of three of them is seven times the mass of the
lightest child. Find the common ratio and the mass of the child whose weight is second
heaviest if the heaviest child is 14.5 kg.
5.2.5
149
CHAPTER 5
Intensive Practice 5.2
Summary OF Chapter 5
�� ��f
��m
�
PROGRESSIONS
�
�a
nth term,
Tn = a + (n – 1)d
Arithmetic
progression
The sum of first n terms,
n
Sn = [2a + (n – 1)d]
2
CHAPTER 5
Solving
problems
Geometric
progression
o�
a��
�
�
�l
a�
The sum to infinity,
S∞ = a , |r| < 1
1–r
�a�
m�
�
f�
nth term,
Tn= arn – 1
The sum of first n terms,
a(rn – 1) , r ≠ 1(|r| > 1)
Sn =
r–1
a(1 – rn), r ≠ 1 (|r| < 1)
Sn =
1–r
WRITE YOUR JOURNAL
Construct a graphical information on the difference between the arithmetic progression and
the geometric progression. Then, think of a situation in your daily life that applies these two
progressions and solve it.
MASTERY PRACTICE
1. –2x – 1, 3x + 2 and 9x + 3 are three consecutive terms of arithmetic progression. Find
(a) common difference,
(b) the first term if 3x + 2 is the third term.
PL1
2. The 9th term of an arithmetic progression is 21 + 3p and the sum of the first three terms is 9p.
Find the common difference. PL2
3. The diagram shows three cylinders such that the volume of each cylinder is arranged
according to arithmetic progression. The sum of the volume of the
first and third cylinder is 24 cm3 and the volume of the fifth
cylinder is 36 cm3. PL3
(a) Find the volume of the smallest cylinder.
(b) Calculate the sum of volume for the first 9 cylinders.
4. The 3rd term of a geometric progression is 30 and the sum of the 3rd and 4th terms is 45.
Find PL2
(a) the first term and the common ratio,
(b) the sum to infinity.
150
Progressions
5. The diagram shows the arrangement of a few chairs. The height of each
chair is 80 cm. When the chairs are arranged, there is a 4 cm gap in
between two chairs. The arranged chairs will be kept in the store. PL4
(a) Find the maximum number of chairs that can be arranged if the height
of the store is 3 m.
(b) 13 chairs will be kept in the store with the condition that the first stack
will have the maximum number of chairs and the arrangement of chairs
for subsequent stacks decreases 2. Calculate the total number of chair
kept in the store.
7. Shahrul has a toy car collection which he collects every month. The number of toy cars
increases every month according to geometric progression. The total number of toy cars in
the first four months is ten times the total number of toy car in the first month. PL5
(a) If r represents the common ratio, show that r4 – 10r2 + 9 = 0. Hence, find the positive
values of r.
(b) Calculate the expenses paid by Shahrul in the 6 months if he started to buy 2 toy cars and
the average price of a toy car is RM7.50.
Exploring
MATHEMATICS
1. Prepare two piggy banks.
2. In 10 days, put money into the piggy bank by following these rules:
First piggy bank:
Start putting 50 cents into the piggy bank on the first day, RM1 on the second day
RM1.50 on the third day and so on. Each day, the total savings increases 50 cents.
Second piggy bank:
Start putting 10 cents into the piggy bank on the first day, 20 cents on the second day
40 cents on the third day and so on. Each day, the total savings increases twice the
amount compared to the previous day.
3. Record the total savings after 10 days.
4. Observe the relationship between the total savings with progression.
5. Prepare a report on the relation between the arithmetic progression and geometric
progression with the total amount of your savings.
151
CHAPTER 5
6. Encik Muslim starts to save RM14 000 into his new born baby’s account. The bank offers a
5% interest yearly. Encik Muslim hopes that his child’s saving will reach RM30 000 when his
child becomes 18 years old. PL4
(a) Do you think he can obtain RM30 000 when his child becomes 18 years old? Show your
calculations.
(b) If the interest decreases to 3% per year after 10 years, calculate the total savings when
Encik Muslim’s child is 18 years old. Can the savings of his child reach RM30 000?
CHAPTER
Linear Law
6
What will be learnt?
CHAPTER 6
Linear and Non-Linear Relations
Linear Law and Non-Linear Relations
Application of Linear Law
List of
Learning
Standards
bit.ly/2VtVKLE
wORDS
KEY W
Linear relation
Non-linear relation
Inspection method
Line of best fit
152
Hubungan linear
Hubungan tak linear
Kaedah pemerinyuan
Garis lurus penyuaian
terbaik
6.1.1
Linear Law
Did you
Know?
Leonardo Bonacci or known as Fibonacci was
a famous mathematician in Italy in the 13th
century. He had discovered a concept such
that the ratio of distance between the tip of
nose and tip of chin and from the cheek to the
tip of chin gave a value which is known as the
Golden Ratio. The ratio of readings can be
measured and represented through a straight
line graph involving two variables.
For further information:
bit.ly/2M4jLFO
OF THIS CH
APTER
In order to relate two variables, a straight line
will help us to get the value of a constant. When
a straight line is drawn from the results of an
experiment, sometimes the data obtained do not
produce a perfect straight line. Therefore, the
data will be represented by a line of best fit.
String art is a type of art which uses
strings or ropes to form geometric
patterns. String art applies the use of
straight lines to form patterns which are
not straight lines.
Scan the QR code to
watch the video showing
the methods of making
string art.
bit.ly/2TWUXAY
6.1.1
153
CHAPTER 6
SIGNIFICANCE
6.1
Linear and Non-Linear Relations
Differentiating linear and non-linear relations
INQUIRY 1
In pairs
Aim: To differentiate between linear and non-linear relations based on tables of data and
graphs
Instructions:
1. Complete the table based on the given equation.
(a) y = 2x2 – 5x + 8
x
–3
–2
–1
0
1
2
3
4
–4
–3
–2
–1
0
1
2
3
y
(b) y = x + 4
x
y
CHAPTER 6
2. Draw the graph of y against x based on the values obtained in both tables for each of
the equations.
3. Based on the graphs drawn, compare the shapes of the graphs for both equations.
What do you observe?
From the results of Inquiry 1, we can conclude that:
FLASHBACK
The graph which forms a straight line is a linear relation whereas
the graph which does not form a straight line is a non-linear
relation.
For a linear graph, Y = mX + c,
X represents the variable
on the horizontal axis, Y
represents the variable
on the vertical axis, m
represents the gradient and
c represents the Y-intercept.
A linear graph can be obtained from a non-linear graph when the
variables of the X-axis or Y-axis or both are changed.
Example 1
Draw the graph of Y against X based on each of the following tables of data and hence determine
which graph is a graph of linear relation? Give your reason.
(a)
X
–3
–2
–1
0
1
2
3
(b)
Y
–17
0
5
4
3
8
25
X
–3
–2
–1
0
1
2
3
Y
154
9
7
5
3
1
–1
–3
6.1.1
Linear Law
Solution
(a)
(b)
Y
Y
Graph of Y against X
–3
–2
30
8 Graph of Y against X
20
6
10
4
–1
0
1
2
3
2
X
–3
–10
–2
0
–1
1
2
3
X
–2
–20
Graph (a) is a graph of non-linear relation because the graph obtained is a curve.
Graph (b) is a graph of linear relation because the graph obtained is a straight line.
1. The diagrams below show two graphs plotted by using the given values in the respective
tables for the equation y = x2 + 1. Which graph shows a graph of linear relation? State your
reasons.
(a)
(b)
x – 6 – 4 –2 0
2
4
6
x2 0
2
4
6
8 10 12
y
37
17
5
1
5
17
37
Graph of y against x
0
7
9
11
13
10
5
10
–2
5
15
20
–4
3
Graph of y against x2
30
–6
1
y
y
40
y
2
4
6
x
0
2
4
6
8
10
12
2. Draw the graph of Y against X based on the given values in the following tables.
(a)
(b)
X
1
3
5
7
9
11
X
2
4
6
10
12
Y
3.16 5.50 9.12 16.22 28.84 46.77
Y
0.5
0.7
0.9
1.3
1.5
x2
14
1.7
Which graph shows a graph of linear relation? State your reasons.
6.1.1
155
CHAPTER 6
Self Practice 6.1
Drawing lines of best fit for graphs of linear relations
A line of best fit is a straight line that joins most of the points plotted on the graph. The points
which are not on the line of best fit must be distributed evenly on both sides of the straight line.
Example 2
The table on the right shows the values
x
2
3
4
6
7
8
9
obtained from an experiment involving
y
0.9 1.1 1.3 1.8 1.9 2.1 2.3
two variables, x and y. Plot the graph of
y against x, by using suitable scales on the x-axis and y-axis. Hence, draw the line of best fit.
Solution
y
Graph of y against x
2.5
2.0
1.5
CHAPTER 6
1.0
0.5
0
2
4
INQUIRY 2
6
In pairs
8
x
10
21st Century Learning
Aim: To draw the line of best fit using digital technology
Instructions:
1. Draw the straight line graph based on the following data
values.
x
1
2
3
4
6
7
y
3
5
6
8
10
11
2. Then, enter the values in the table provided into the
Desmos software by using the same data values as given in
the table above.
3. Follow the diagrammatic steps to draw the line of best fit by
scanning the QR code on the right.
4. Compare the line of best fit obtained in the Desmos software
with the graph drawn.
156
QR
Steps to draw line of
best fit using Desmos
application.
bit.ly/33jaKi3
6.1.2
Linear Law
From the results of Inquiry 2, it is observed that:
The straight line obtained from the graph drawn is the same as the straight line drawn using
the Desmos software. The line is the line of best fit.
Self Practice 6.2
1. The following table shows the values obtained from an experiment involving two variables, x
and y.
x
y
5
8
10
14.5
15
18
20
23
25
26.5
30
33
Plot the graph of y against x, by using suitable scales on the x-axis and y-axis. Hence, draw
the line of best fit.
2. An experiment was carried out to determine the relation between extension of spring, L and
mass of load, m, which was hung at the end of the spring. The following table shows the
results from the experiment.
m (g)
L (cm)
20
0.65
40
1.25
60
1.80
80
2.40
100
2.95
120
3.55
Forming equations of lines of best fit
The equation of straight lines can be written in the form of Y = mX + c, if the gradient, m and
Y-intercept, c are known or can be determined by using any two points on the straight line.
Example 3
The graph below shows part of a straight line obtained by plotting 1 against 1 . Express y in
y
x
terms of x.
1
––
y
20
B(19, 20)
15
10
5 A(1, 7)
0
6.1.2
6.1.3
5
10
15
20
1
––
x
157
CHAPTER 6
Plot the graph of L against m, by using suitable scales on the m-axis and L-axis. Hence, draw
the line of best fit.
Solution
Gradient, m = 20 – 7
19 – 1
= 13
18
Y = mX + c
1 =m 1 +c
y
x
13
(19) + c
20 =
18
Y-intercept
c = 113
18
()
( )
()
1 = 13 1 + 113
y 18 x
18
18x
y=
13 + 113x
Thus,
Equation of straight line
Example 4
CHAPTER 6
The following table shows the experimental values of two variables, x and y.
x
y
1
1.5
3.4
4.2
2
4.8
2.5
5.8
3
6.2
3.5
6.6
4
7.6
(a) Plot the graph of y against x, by using a scale of 1 cm to 1 unit on the x-axis and 1 cm to
2 units on the y-axis. Hence, draw the line of best fit.
(b) From the graph, find the y-intercept and gradient of the line of best fit.
(c) Determine the equation of the line of best fit.
Solution
(a)
(b) From the graph, y-intercept, c = 2
gradient, m = 7.6 – 2
4–0
= 1.4
(c) The equation of line of best fit is
y = 1.4x + 2.
y
8
Graph of y against x
6
4
2
0
158
1
2
3
4
x
6.1.3
Linear Law
t
Self Practice 6.3
1.2
1. The graph of line of best fit in the diagram on the right
shows the values obtained from an experiment which
involves two variables, x and t. Express t in terms of x.
Graph of t against x
1.0
0.8
0.6
0.4
0.2
0
2. The following table shows the experimental values of
two variables, x and y.
x
y
10
16.5
20
20.0
30
23.5
40
27.5
0.1 0.2 0.3 0.4 0.5 0.6
50
31.5
x
60
35.0
(a) Plot the graph of y against x, by using a scale of 2 cm to 10 units on the x-axis and 2 cm
to 5 units on the y-axis. Hence, draw the line of best fit.
(b) From the graph, find the y-intercept and gradient of the line of best fit.
(c) Determine the equation of the line of best fit.
Based on the line of best fit, you can predict the values of variables x or y which are not in the
experiment without repeating the experiment. If the values of variable x or y are outside the
range of points, you can find the value of the variable by extrapolating the drawn straight line
or it can be determined by forming the equation of the straight line.
Example 5
The following table shows the data of two variables, x and log10 y, obtained from an
experiment.
x
log10 y
0.1
0.40
0.3
0.80
0.4
1.04
0.7
1.60
0.8
1.78
1.0
2.20
(a) Plot log10 y against x, by using a scale of 1 cm to 0.2 unit on the x-axis and 1 cm to 0.5 unit
on the log10 y-axis. Hence, draw the line of best fit.
(b) From the graph, find the value of
(i) log10 y when x = 0.9,
(ii) y when x = 0,
(iii) x when log10 y = 2,
(iv) x when y = 20.
(c) Find the equation of the line of best fit.
6.1.3
6.1.4
159
CHAPTER 6
Interpreting information based on the lines of best fit
Solution
(a) log10 y
MATHEMATICS
POCKET
Graph of log10 y against x
• Gradient is the rate of
change of a variable
with respect to another
variable.
• Y-intercept is the
y-coordinate of the
intersection point of a
straight line with the
y-axis.
2.0
1.5
1.0
0.5
CHAPTER 6
0
0.2
0.4
0.6
0.8
x
1.0
(b) (i) From the graph, when x = 0.9, log10 y = 2.
(ii) From the graph, when x = 0,
log10 y = 0.2
y = 100.2
y = 1.585
(iii) From the graph, when log10 y = 2, x = 0.9.
(iv) From the graph, when y = 20, log10 20 = 1.3.
Then, x = 0.56.
(c) Two points are selected from the graph, they are (0.7, 1.60)
and (0.3, 0.80).
Gradient, m = 1.60 – 0.80
0.7 – 0.3
=2
Y-intercept, is 0.2.
Thus, the equation of the line of best fit is log10 y = 2x + 0.2.
Self Practice 6.4
1. The following table shows the values of x and y obtained from an experiment.
x
1
2
4
6
8
10
15
y
5.5
7.0
10.5
13.0
15.5
19.0
26.5
(a) Plot y against x, by using a scale of 2 cm to 2 units on the x-axis and 2 cm to 5 units on
the y-axis. Hence, draw a line of best fit.
(b) From the graph, find
(i) the y-intercept,
(ii) the value of y when x = 12,
(iii) the gradient,
(iv) the value of x when y = 15,
(c) Find the equation of the line of best fit. Hence, calculate the value of y when x = 28.
160
6.1.4
Linear Law
Intensive Practice 6.1
Scan the QR code or visit at bit.ly/2pbnnNb for the quiz
1. The following tables show the experimental data involving variables x and y.
(a)
x
–4
–2
–1
0
1
2
(b)
y
3
–3
–3
–1
3
9
1
x
0.80
0.70
0.50
0.40
0.25
0.20
4.00
4.41
5.20
5.62
6.20
6.40
y2
Draw the graphs based on the data in the tables. Then, determine the graph which shows
linear relation and non-linear relation. Give a reason for your answer.
2. Based on an experiment, the values of X and of Y are related as in the following table.
X
Y
20
108.0
30
110.4
40
112.4
50
114.4
60
116.8
70
119.0
Plot the graph of Y against X and draw the line of best fit. Then, write the equation of the line
of best fit.
log10 (x + 1)
0.18
log10 y
0.33
0.30
0.45
0.50
0.64
0.60
0.75
0.70
0.85
0.78
0.93
(a) Plot the graph of log10 y against log10 (x + 1), by using a scale of 2 cm to 0.1 unit on the
log10 (x + 1)-axis and log10 y-axis. Hence, draw the line of best fit.
(b) From the graph, find
(i) the gradient,
(ii) the log10 y-intercept,
(iii) the value of x when log10 y = 0.55,
(c) Calculate
(i) the value of y when x = 2.5,
(ii) the value of x when y = 1.5.
4. The results of experiment of two variables, x2 and xy, are shown in the following table.
x2
xy
5
12
9
15.5
16
22
25
30
36
40
42
45
(a) Plot the graph of xy against x2, by using a scale of 2 cm to 5 units on X-axis and Y-axis.
Hence, draw the line of best fit.
(b) From the graph, find
(i) the gradient,
(ii) the Y-intercept,
(iv) the value of y when x = 2.5.
(iii) the value of x2 when xy = 16.5,
(c) Calculate the value of x when xy = 100.
161
CHAPTER 6
3. The following table shows the readings of two variables, log10(x + 1) and log10 y.
6.2
Linear Law and Non-Linear Relations
Applying linear law to non-linear relations
By using linear law, most of the non-linear relations can be converted to linear relations so that
a straight line can be drawn. It is easier to obtain information from the straight line graphs than
from the curves.
The non-linear equation y = ax + b , such that a and b are constants can be converted to linear
x
equation form Y = mX + c by using two methods.
Smart TIPS
Method 1
y = ax + b
x
Multiply both sides of the equation by x
y(x) = ax(x) + b (x)
x
yx = ax2 + b
Compare with Y = mX + c
xy = ax2 + b
Through comparison, Y = xy, X = x2, m = a and c = b.
Y
xy
m
a
X
x2
c
b
CHAPTER 6
Method 2
y = ax + b
x
y b ax
= +
x x2 x
y 1
= (b) + a
x x2
Divide both sides of the equation by x
Compare with Y = mX + c
Through comparison, Y =
m = b and c = a.
You have to choose suitable
variables for X and Y to
change the non-linear
equation to the linear form,
Y = mX + c such that m is
the gradient of the straight
line and c is the y-intercept.
The variables X and Y must
contain variables only and
they cannot contain the
unknown constants. m and c
must contain only constants.
y
, X = 12,
x
x
Y
m
y
x
b
X
1
x2
c
a
Example 6
Convert the equation y = pqx such that p and q are constants to the linear form Y = mX + c.
Hence, identify Y, X, m and c.
Solution
y = pqx
log10 y = log10 p + x log10q
log10 y = log10q (x) + log10 p
Write both sides of the equation in logarithmic form
Compare with Y = mX + c
Through comparison, Y = log10 y, X = x,
m = log10q dan c = log10 p
162
Y
log10 y
m
log10q
X
x
c
log10 p
6.2.1
Linear Law
Example 7
The table below shows the values of x and y obtained from an experiment. The variables, x
and y are related by the equation 3y – px2 = qx, such that p and q are constants.
x
y
1
20
2
34
3
48
5
60
7
63
9
36
(a) Convert the equation 3y – px2 = qx to the linear form.
y
(b) Plot the graph of against x, by using a scale of 1 cm to 2 units on the x-axis and 1 cm to
x
y
5 units on the -axis. Hence, draw the line of best fit.
x
(c) From the graph, find the value of p and of q.
Solution
y
x
y
–x
20
20
17
16
12
9
Graph of y–x against x
4
Mathematics Museum
Renè Descartes invented
the coordinate grids known
as Cartesian Diagram. How
did the idea of inventing
Cartesian Diagram trigger
him? He laid down on his
bed until late night and
observed a housefly on
the ceiling of his room. He
thought of the best way
to illustrate the position of
the housefly on the ceiling.
He decided to take one
corner of the ceiling as the
reference point.
For further information:
15
10
bit.ly/2oALd54
5
0
2
4
(c) From the graph,
y-intercept = 22
q
= 22
3
q = 66
6.2.1
6
8
10
x
gradient = 4 – 22
9–0
= – 2
p
= –2
3
p = –6
163
CHAPTER 6
(a)3y – px2 = qx
Divide both sides of the
3y px2 qx
=
–
equation by 3x
3x 3x 3x
y px q
–
=
x
3 3
y p
q
Compare with Y = mX + c
= (x) +
x 3
3
y
p
q
Through comparison, Y = , X = x, m = and c = .
x
3
3
(b)
x
1
2
3
5
7
9
Example 8
The table below shows the values of x and y obtained from an experiment. The variables x and
px
y are related by the equation y = , such that p and q are constants.
q
x
2
4
5
6
7
8
10
y
0.3162
5.0119
100
1 584.89
6 309.57
63 095.73
100 000
(a) Plot the graph of log10 y against x, by using a scale of 1 cm to 2 units on both the log10
y-axis and x-axis. Hence, draw the line of best fit.
(b) From the graph, find
(i) the value of p and of q,
(ii) the value of y when x = 3.
Solution
px
q
log10 y = log10 px – log10 q
log10 y = x log10 p – log10 q
log10 y = (log10 p)x – log10 q
(a)
y=
Compare with Y = mX + c
CHAPTER 6
Through comparison, Y = log10 y, X = x, m = log10 p and c = – log10 q
x
log10 y
2
– 0.50
4
0.70
5
2.00
6
3.20
7
3.80
8
4.80
10
5.00
4
(b) (i) – log10 q = –2.5
log10 q = 2.5
q = 316.228
log10 p = 2.00 – 3.80
5–7
log10 p = 0.9
p = 7.943
2
(ii) When x = 3, log10 y = 0.2
y = 1.585
log10 y
Graph of log10 y against x
6
0
2
4
6
8
10
x
–2
164
6.2.1
Linear Law
Self Practice 6.5
1. Convert the following non-linear equations into the form Y = mX + c. Hence, identify Y, X, m
and c.
p
(b) y = hx2 + x
(c) y = 2 + q
(a) y = px2 – q
x
2. The table below shows the values of x and y obtained from an experiment. The variables !w
x
1
1
and are related by the equation = p!w
x + q, such that p and q are constants.
y
y
!w
x
1
y
0.70
1.00
1.22
1.45
1.58
1.80
0.62
1.20
1.65
2.00
2.38
2.75
(a) Plot the graph of 1 against !w
x , by using a scale of 1 cm to 0.5 unit on both the !w
x-axis
y
and 1 -axis. Hence, draw the line of best fit.
y
(b) From the graph, find the value of
(i) q,
(ii) p,
(iii) y when x = 1.21.
Intensive Practice 6.2
1. Convert the following non-linear equations to the linear form. Hence, identify Y, X, gradient
and Y-intercept.
q
(a) y = 5x2 + 3x
(b) y = p!w
x+
(c) y = axb
!w
x
(d) x = mxy + ny
(e) ypx = q
(f) y(b – x) = ax
2. The table below shows the data which relates the variables x and y by the equation y = ax3 +
bx2, such that a and b are constants.
x
y
0.5
0.31
1.0
2.05
1.5
6.19
2.0
14.00
2.5
26.30
3.0
45.00
(a) Convert the non-linear equation y = ax3 + bx2 to the linear form.
y
y
(b) Plot the graph of 2 against x, by using suitable scales on the x-axis and 2 -axis. Hence,
x
x
draw the line of best fit.
(c) From the graph, find the value of a and of b.
3. The table below shows the data which relates the variables x and y by the equation y = ab + x,
such that a and b are constants.
x
y
1
2.83
2
5.66
3
11.31
4
22.63
5
45.25
(a) Convert the non-linear equation y = ab + x into a linear equation.
(b) Plot the graph of log10 y against x, by using suitable scales on the x-axis and log10 y-axis.
Hence, draw the line of best fit.
(c) From the graph, find the value of a and of b.
6.2.1
165
CHAPTER 6
Scan the QR code or visit bit.ly/2M54EMF for the quiz
6.3
Application of Linear Law
Solving problems involving linear law
Example 9
MATHEMATICS APPLICATION
An experiment is carried out to study the effect of the growth of a plant on the concentration
of a hormone. The readings from the experiment are recorded in the table below. The growth
of the plant and the concentration of hormone are related by the equation P = 180 + rK – sK 2,
such that r and s are constants.
Concentration of hormone per million (K)
% growth of plant (P)
1
181
3
179.7
4
178
6
168
8
157
10
140
(a) Plot the graph of P – 180 against K, by using a scale of 2 cm to 2 units on the X-axis and
K
2 cm to 1 unit on the Y-axis. Hence, draw the line of best fit.
(b) From the graph, calculate the value of r and s.
Solution
CHAPTER 6
1 . Understanding the problem
◆
◆
◆
Identify the variables to determine the X-axis and Y-axis.
Plot the graph by using the scales given.
Based on the graph, find the value of r and s.
2 . Planning the strategy
(a) Convert the non-linear equation to the linear form and compare with the form
Y = mX + c, such that m is the gradient and c is the Y-intercept.
(b) Construct a new table using the new variables.
(c) Plot the graph by using the values in the new table.
(d) Find the y-intercept and gradient by refering to the graph. Hence, compare with
the equation Y = mX + c.
3 . Implementing the strategy
(a)
P = 180 + rK – sK2
P – 180 = rK – sK2
P – 180 = rK – sK2
K
K
K
P – 180 = r – sK
K
P – 180 = –sK + r
K
166
6.3.1
Linear Law
Through comparison, Y = P – 180, X = K, m = –s and c = r.
K
K
P – 180
K
1
3
4
6
8
10
1.00
− 0.10
− 0.50
−2.00
−2.88
− 4.00
P
– 180
K
2
(b) Y-intercept = 1.6
r = 1.6
– 180 against K
Graph of P
K
1
0
2
4
6
8
10
K
Gradient, –s = – 4 – 1.6
10 – 0
–s = – 0.56
s = 0.56
–1
–2
–3
–4
CHAPTER 6
4 . Making a conclusion
When K = 1,
When K = 3,
P = 180 + rK – sK2
P = 180 + rK – sK2
= 180 + (1.6)(1) – (0.56)(1)2
= 180 + (1.6)(3) – (0.56)(3)2
= 181.04 = 179.76
≈ 181 ≈ 179.7
Self Practice 6.6
1. The table below shows the total population of a type of bacteria in a test tube. The variable
x represents the number of hours and y represents the total population. Variables x and y are
related by the equation y = pqx, such that p and q are constants.
x (Number of hours)
y (Total population)
2
3.98
4
6.31
6
10.00
8
15.85
10
25.12
16
100.00
(a) Plot log10 y against x, by using suitable scales on both axes. Hence, draw the line of best
fit.
(b) From the graph, find the value of
(i) p
(ii) q
(c) Estimate the total population of bacteria after 5 hours.
6.3.1
167
2. The table below shows the values of two variables, x and y obtained from an experiment. The
variables x and y are related by the equation xy – yb = a, such that a and b are constants.
x
y
0.485
1.556
20.60
4.528
18.00
10.227
13.25
18.333
8.80
6.00
100.000
1.40
(a) Plot y against xy, by using suitable scales on both axes. Hence, draw the line of best fit.
(b) From the graph, find the value of a and of b.
(c) Another method of getting a straight line graph for the above non-linear equation is by
plotting 1 against x. Without drawing the second graph, calculate the gradient and the
y
Y-intercept of the graph.
Intensive Practice 6.3
Scan the QR code or visit bit.ly/2p2uoA1 for the quiz
1. Diagram (a) and Diagram (b) show two straight line graphs which are related by the equation
y!w
x = 10. State the value of p in the following cases.
(a)
(b)
y
y
x
(2, p)
CHAPTER 6
p
x
0
1
x
0
Diagram (a)
Diagram (b)
2. The table below shows the data obtained from an experiment on a pendulum such that p is the
length, in cm of the pendulum and t is the period of oscillations, in seconds. One of the values
of t was recorded wrongly.
Length, p (cm)
Period of oscillations, t (s)
10
6.3
20
9.0
30
11.0
40
12.6
50
14.1
60
15.0
(a) Plot the graph of t2 against p, by using suitable scales. Hence, draw the line of best fit.
(b) Mark ⊗ on the graph, for the point which was recorded wrongly. Then, find the correct
value of t.
p = t , such that
(c) Use the graph to find the value of k if t and p are related by the equation !w
k
t and p are constants.
168
6.3.1
Linear Law
3. The total production of a type of commodity, N, is related to the total number of hours, H by
the equation 2N 2 – a = b . The table below shows the corresponding value of N and of H.
H
H (hours)
(a)
(b)
(c)
(d)
N (metric tonnes)
20
1.225
40
60
1.162
80
1.140
100
1.135
1.127
Plot the line of best fit of N 2H against H, by using suitable scales.
Use the graph in (a) to find the value of a and of b.
From the graph, estimate the total production if the total number of hours is 10.
The manager of the company plans to produce1.1183 metric tonnes of commodity. If a
worker works for 8 hours, how many workers are needed by the company?
4. The table below shows the values in an experiment involving the concentration of liquid, L
b
unit3, which is related to temperature, T, by the equation L = A(3) T .
3
L (unit )
0.100
6.31 × 108
0.033
0.020
0.014
0.011
0.010
1.00 × 1010 1.58 × 1011 3.98 × 1012 2.51 × 1013 1.58 × 1014
(a) Plot the line of best fit of log10 L against 1 , by using suitable scales.
T
(b) Use the graph in (a) to find the value of
(i) A,
(ii) b.
(c) Determine the temperature when the liquid is heated until its concentration achieves 21.5
unit3.
5. The table below shows the points obtained in an experiment involving two variables, u and v
which are related by the relation 1 + 1 = 1 .
v u f
u
v
15
30.0
20
20.2
25
16.6
50
12.5
100
11.1
(a) Plot 1 against 1 . Draw the line of best fit.
v
u
(b) From the graph,
(i) express v in terms of u.
(ii) determine the value of 1 when 1 = 0, Hence, find the value of f.
f
u
169
CHAPTER 6
T (°C)
Summary OF Chapter 6
i��o��i��
Solve problems
involving linear law
Conversion of
non-linear
equation to
linear form
LINEAR
LAW
�� ��r��
Linear relations
Graph of line
of base fit
a��o�
�l��
�
a
Equation
Y = mX + c
��f � � m �
�
g
�� ��e��i�
�s
Non-linear
relations
CHAPTER 6
Interpret information
from graph of line
of best fit
WRITE YOUR JOURNAL
dismantle
assemble
The above diagram shows the building blocks which are assembled and can be dismantled. In
mathematics, there are many examples with inverses. You can convert non-linear equations into
linear equations and vice versa. Can you determine the steps needed to convert linear equations
into non-linear equations?
170
Linear Law
MASTERY PRACTICE
1. Express the following non-linear equations in the linear form, Y = mX + c, such that X and Y
are variables, m and c are constants. PL2
p q
(a) y = 3x + 42
(b) y = px3 + qx2
(c) y = + x
x
x p
x
k
(e) y = pk x – 1
(f) y =
(d) y = pk !wx
p
2
2. The variables x and y are related by the equation y = px2 + qx,
such that p and q are constants. The diagram on the right shows
part of the line of best fit obtained by plotting the graph
y
of against x. PL3
x
(a) Convert the equation y = px2 + qx to the linear form.
(b) Find the value of p and q.
y
–
x
(1, 5)
(9, 3)
x
0
log10y
x2
3. The variables x and y are related by the equation y = pq 4 .
The diagram on the right shows part of the line of best fit
obtained by plotting log10 y against x2. Find the value of p
and q. PL3
(6, 5)
(4, 4)
x2
4. The diagram on the right shows part of the line of best fit of
against x. Given y = 5x – 3x2, find the value of k and h.
CHAPTER 6
0
y
–x
y
x
(2, k)
PL3
(h, 3)
x
0
5. The diagram on the right shows part of the graph of y against x
for the equation y = abx, such that a and b are constants. PL3
(a) Sketch the straight line graph of log10 y against x. Mark and
state the coordinates of the corresponding point P and Q.
(b) Based on the graph in (a), find the value of a and b.
6. When x2y against x are plotted, a straight line is obtained. The
straight line has a gradient of 8 and it passes through point (2, 19).
(a) Determine the equation which relates x and y.
(b) Hence, find the value of y when x = 9.4.
y
Q(3, 32)
0
P(1, 4)
x
171
7. A study is conducted to determine the relation between the mass, m and the volume, V of a
type of cooking oil. The following table shows the results of the study. PL2
V
0.5
m
1.0
0.35
1.5
0.84
2.0
1.23
2.5
1.60
3.0
2.00
2.37
Plot the graph of m against V by using a scale of 2 cm to 1 unit on both axes. Hence, draw the
line of best fit.
8. Based on an experiment, the relation between the values of x and the values of y are shown in
the table below. PL3
x
10
y
20
16.5
30
20.0
40
23.5
50
27.5
60
31.5
35.0
(a) Plot the graph of y against x and draw the line of best fit by using a scale of 2 cm to
10 units on the x-axis and 2 cm to 5 units on the y-axis.
(b) Hence, form the equation of the straight line.
9. The table below shows the values which relate the temperature, T of a solution after time, t in
an experiment. PL4
t(s)
2
CHAPTER 6
T(°C)
4
29.0
6
40.0
8
31.0
10
32.1
33.0
(a) Plot the graph of T against t. Hence, draw the line of best fit by using suitable scales.
(b) Mark ⊗ on the graph the point which was recorded wrongly. Then, find the correct value
for T °C.
(c) From the graph, find
(i) the initial temperature of the solution,
(ii) the temperature of the solution after 9 seconds,
(iii) the time taken for the solution to reach a temperature of 30.5°C.
10. The table below shows the values of two variables, x and y, obtained from an experiment. The
variables x and y are related by the equation y = st x, such that s and t are constants. PL3
x
y
1.5
2.51
3.0
3.24
4.5
4.37
6.0
5.75
7.5
7.76
9.0
10.00
(a) Plot the graph of log10 y against x, by using a scale of 2 cm to 1 unit on the x-axis and
2 cm to 0.1 unit on the log10 y-axis. Hence, draw the line of best fit.
(b) From the graph, find the value of
(i) s,
(ii) t,
(iii) x when y = 4.
172
Linear Law
11. The table below shows the values of two variables, x and y, obtained from an experiment. The
q
variables x and y are related by the equation 2y – p = , such that p and q are constants. PL3
x
x
y
1
2
5
3
3.5
4
3.1
2.7
5
2.6
6
2.5
(a) Plot the graph of xy against x, by using a scale of 2 cm to 1 unit on the x-axis and 2 cm to
2 units on the xy-axis. Hence, draw the line of best fit.
(b) Use the graph in (a) to find the value of
(i) p,
(ii) q,
(iii) y when x = 3.5.
(c) Calculate the value of x when y = 50.
Exploring
MATHEMATICS
Age (year)
Mass (kg)
1
0.5
2
1.0
3
2.0
4
2.8
CHAPTER 6
Durian is a well-known fruit in Southeast Asia. Attempts to
export durians in frozen form have been carried out in order to
promote durians to countries outside Southeast Asia. Manuring
of plants must be carried out regularly in order to increase the
production of durians. The following table shows the relation
between the age and the mass of durian trees by using the
recommended manuring method during the vegetative stage.
5
4.0
(a) Draw a dispersion diagram for the data in the table. Does the dispersion diagram show a
linear relation between the age and mass of durian trees which used the manuring method?
(b) By using suitable scales, draw the line of best fit by taking the mass as a dependent
variable and the age as an independent variable. Then, find the equation relating the two
variables.
(c) Convert the non-linear relation to the linear form and construct a new table for the
variables involved.
(d) From your graph, predict the mass of a durian tree of 7 years.
173
Coordinate
Geometry
CHAPTER
7
What will be learnt?
Divisor of a Line Segment
Parallel Lines and Perpendicular Lines
Areas of Polygons
Equations of Loci
CHAPTER 7
List of
Learning
Standards
bit.ly/2p7Q7qq
wORDS
KEY W
Divisor of line
segment
Parallel straight lines
Perpendicular
straight lines
Gradient
Area of polygon
Equation of a locus
174
Pembahagi tembereng
garis
Garis lurus selari
Garis lurus serenjang
Kecerunan
Luas poligon
Persamaan lokus
7.1.1
Coordinate Geometry
Did you
Know?
Ibrahim Ibn Sinan (908 – 946 AD) is a
mathematician and astronomer from Harran
in northern Mesopotamia. He started doing
research on geometry and astronomy at the
age of 15 and recorded his first research
results at the age of 16. He continued
Archimedes’ research on area, volume and in
particular, tangent to a circle.
For further information:
bit.ly/2B3TfpM
SIGNIFICANCE
APTER
In construction, coordinate geometry is used
when drawing sketches of buildings.
Astrophysicists use coordinate geometry to
determine the distance between planets.
Coordinate geometry is used in aviation to
determine the angles in the flight paths of
aircraft.
The use of a GPS (Global Positioning
System) navigation application allows
us to quickly and easily locate the
places that we want to go. Do you
know that GPS navigation uses the
idea of coordinate geometry known
as World Geodetic System (WGS 84)
to determine the location of places on the
surface of the Earth?
7.1.1
Scan this QR code
to watch video on
application of GPS.
bit.ly/2JyqSoC
175
CHAPTER 7
OF THIS CH
7.1
Divisor of a Line Segment
A line segment is part of a straight line with two end points with specific length or distance.
Any point dividing the segment in a particular ratio is known as the internal point.
Relating the position of a point that divides a line segment with the
related ratio
CHAPTER 7
INQUIRY 1
In groups
Aim: To explore the relationship between the position of a
point on a line segment and its ratio
Instructions:
1. Scan the QR code or visit the link on the right.
bit.ly/2IvAJe2
2. Drag point P on the line segment AB to the left and right.
Observe the values of m and n displayed.
3. What is the relationship between the position of point P on the line segment and the
values of m and n?
4. Consider the position of point P on the line segment AB and answer the following
questions.
(a) How many parts are there between point P and point A?
(b) How many parts are there between point P and point B?
(c) How many parts are there between point A and point B?
(d) What are the lengths of AP and PB in terms of AB?
(e) Determine the ratio AP : PB.
(f) What is the relationship between the position of point P on line segment AB and the
ratio obtained in (e)?
5. Now, drag point P so that the ratio m : n is 5 : 5 = 1 : 1. Is the length of AP same as PB?
Determine the position of point P on the line segment when the ratio m : n is the same
for every part.
6. Change the ratio m : n and observe the position of point P. Does the position change
with the change in ratio value?
From the results of Inquiry 1, point P on the line segment AB
divides the line segment into two parts in the ratio m : n. Ratio
m : n means line segment AB is divided into (m + n) equal parts.
m
A
n
●
P
B
The position of point P on line segment AB determines the m number of equal parts from
point A to point P and the n number of equal parts from point B to point P. So, point P divides
the line segment in the ratio m : n. Conversely, the ratio m : n will determine the position of point
P on the line segment AB. When ratio m : n changes, the position of point P will also change. If
m = n, then point P is the midpoint of the line segment AB. In general,
The position of point P on a line segment AB divides the line segment
in the ratio m : n and vice versa.
176
7.1.1
Coordinate Geometry
Example 1
7
Given line segment PQ, a point R is on PQ. Point R is of the distance PQ from point P
9
along the line segment PQ.
(a) Sketch this situation by using the line segment.
(b) Is point R closer to P or Q? Explain.
(c) By using the information given, determine the following ratios.
(i) PR : PQ,
(ii) RQ : PR,
(iii) PR : RQ.
(d) Hence, describe the relationship between the position of point R on line segment PQ and
its ratio.
Solution
R
(a) P
Q
●
(b) Point R is closer to Q because the position of point R is more than half of the line segment
from point P.
(c) (i) PR : PQ = 7 : 9
(ii) RQ : PR = 2 : 7
(iii) PR : RQ = 7 : 2
(d) Point R divides the line segment PQ in the ratio 7 : 2
7
P
2
Q
●
R
1. The diagram below shows a line segment AB that is divided into 12 equal parts.
P
A
●
Q
R
●
●
B
P, Q and R are internal points on the line segment.
(a) Determine the position of each point in relation to its ratio.
(b) If point S is on the line segment AB in the ratio 5 : 7, mark and label the position of point
S on the line segment.
2. The diagram below shows point P which divides a piece of rope AB in the ratio m : n.
A
m
n
P
B
Given AP = 10 cm and AB = 35 cm.
(a) Find the values of m and n.
(b) Describe the position of P on the rope in relation to its ratio.
(c) If the rope is placed on the x-axis of the Cartesian plane such that A is the origin and
coordinates of B is (21, 0), determine the coordinates of P.
7.1.1
177
CHAPTER 7
Self Practice 7.1
Deriving the formula for divisor of a line segment on a Cartesian plane
In Diagram 7.1, the coordinates of points A and B are
(x1, y1) and (x2, y2) respectively. P(x, y) is a point which
divides line segment AB in the ratio m : n. So, CD = x – x1,
DE = x2 – x, PG = y – y1 and BF = y2 – y. Since AC, PD and
BE are parallel, we get:
CHAPTER 7
CD AP
=
DE PB
x – x1 m
x2 – x = n
n(x – x1) = m(x2 – x)
nx – nx1 = mx2 – mx
mx + nx = nx1 + mx2
x(m + n) = nx1 + mx2
nx1 + mx2
x= m
+n
y
B(x2, y2)
n
P(x, y)
F
m
A(x1, y1)
0
For BF, PG and AC which are also parallel, we get:
PG AP
=
BF PB
y – y1 m
y2 – y = n
n(y – y1) = m(y2 – y)
ny – ny1 = my2 – my
my + ny = ny1 + my2
y(m + n) = ny1 + my2
ny1 + my2
y= m
+n
G
x
C
D E
Diagram 7.1
QR
Scan the QR code below
for another method of
deriving formula for
divisor of line segment
on a Cartesian plane.
bit.ly/35ml4aS
Thus, the coordinates of point P(x, y) which divides the line
segment joining points A(x1, y1) and B(x2, y2) in the ratio m : n is:
P(x, y) =
(nxm++mxn , nym++myn )
1
2
1
2
What will happen when m = n? When m = n, P will become
the midpoint of line segment AB and is represented by M.
(mxm ++ mmx , mym ++ mmy )
m(x + x ) m(y + y )
=(
,
2m
2m )
x +x y +y
=(
,
2
2 )
M=
1
2
1
1
178
1
2
2
1
2
1
2
2
BRAINSTORMING
Using Pythagoras Theorem,
show that the distance
of line segment AB
represented by d in
Diagram 7.1 is
d = !(x2 – x1)2 + (y2 – y1)2
7.1.2
Coordinate Geometry
Example 2
(a) The coordinates of points A and B are (–5, –2) and (5, 8). If point P divides line segment
AB in the ratio 3 : 2, find the coordinates of point P.
(b) Points A(–7, 3), P(5, –3), B and M are on a straight line. Given P divides line segment AB
in the ratio 3 : 1 and M is the midpoint of AB. Find
(i) the coordinates of B,
(ii) the coordinates of M.
Solution
(b) (i) B is (x, y) and P(5, –3) divides AB
in the ratio 3 : 1. So,
x-coordinate of P = 5
1(–7) + 3x
=5
3+1
3x – 7 = 20
3x = 27
= 9
y-coordinate of P = –3
1(3) + 3y
= –3
3+1
3 + 3y = –12
3y = –15
y = –5
Thus, coordinates of point B are (9, –5).
(x +2 x , y +2 y )
–7 + 9 3 + (–5)
,
=(
2
2 )
(ii) Midpoint of AB =
1
2
1
2
= (1, –1)
Thus, coordinates of point M are (1, –1).
7.1.2
y
B(5, 8)
2
P(x, y)
3
x
0
A(–5, –2)
y
A(–7, 3)
3
0
P(5, –3)
1
x
B(x, y)
y
A(–7, 3)
1
0
M
1
x
B(9, –5)
179
CHAPTER 7
(a) P(x, y) is the point that divides AB in the ratio 3 : 2. So,
2(–5) + 3(5)
x-coordinate of P, x =
3+2
–10 + 15
=
5
5
=
5
=1
2(–2) + 3(8)
y-coordinate of P, y =
3+2
– 4 + 24
=
5
20
=
5
=4
Thus, coordinates of point P are (1, 4).
Example 3
Find the ratio AP : PB such that point P(–1, 2) divides the line segment joining point A(–2, 1)
and point B(2, 5).
Solution
Let P(–1, 2) divides AB in the ratio m : n and the x-coordinate of P is –1.
y
n(–2) + m(2)
= –1
B(2, 5)
m+n
2m – 2n = –m – n
n
3m = n
P(–1, 2)
m 1
=
m
n 3
Thus, ratio AP : PB is 1 : 3.
A(–2, 1)
0
x
Self Practice 7.2
1. Point P divides the line segment joining points A and B below in the given ratios. Find the
coordinates of point P.
(a) A(3, 7), B(–7, 2) in the ratio 3 : 2.
(b) A(– 4, –1), B(2, 5) in the ratio 2AP : PB.
(c) A(7, –3), B(–3, 2) in the ratio 3AP : 2PB.
CHAPTER 7
2. Point R( p, t) divides the line segment joining points A(2h, h) and B(2p, 3t) in the ratio 2 : 3.
Express p in terms of t.
3. A straight line passes through points A(−2, −5) and B(6, 7). Point C divides the line segment
AB in the ratio 3 : 1 while D divides AB in the ratio 1 : 1. Find
(a) the coordinates of C,
(b) the coordinates of D.
4. Point P divides the line segment joining the points A and B in the ratio AP : PB. Find the ratio
AP : PB and the value of k for each of the following.
(b) A(1, 2), B(k, 6) and P(3, 4)
(a) A(1, k), B(–5, 10) and P(–1, 2)
(d) A(–3, –2), B(2, 8) and P(–1, k)
(c) A(k, 3), B(2, 8) and P(6, 4)
Solving problems involving divisor of line segment
Example 4
Mathematics
Application
A spider is at position E(–7, –5) on a graph paper and moves towards point G(13, 5) along a
straight line with uniform velocity. The spider is at point P after moving for 18 seconds and
arrives at point G in 1 minute. Determine
(a) the coordinates of point P,
(b) the ratio EQ : QG when the spider is at point Q(11, 4) on the straight line.
180
7.1.2
7.1.3
Coordinate Geometry
Solution
1 . Understanding the problem
The original position of the spider is E(–7, –5). The spider arrives at point G(13, 5) in 1
minute (60 seconds).
◆ Find the coordinates of P after moving for 18 seconds.
◆ Find the ratio EQ : QG when the spider is at point Q(11, 4).
◆
2 . Planning the strategy
◆
Find the ratio EP : PG first and use the formula for divisor of line segment,
nx1 + mx2 ny1 + my2
P(x, y) = m
+ n , m + n to determine the coordinates of P.
(
)
Use the formula
(nxm++mxn , nym++myn ) again to determine the ratio EQ : QG.
1
2
1
2
3 . Implementing the strategy
(a) The spider is at P(x, y) after moving for 18 seconds.
The ratio EP : EG is 18 : 60 = 3 : 10, so the ratio
EP : PG = 3 : 7.
nx1 + mx2 ny1 + my2
P(x, y) = m
+n , m+n
7(–7) + 3(13) 7(–5) + 3(5)
=
,
3+7
3+7
–10 –20
=
,
10 10
= (–1, –2)
Thus, the coordinates of P are (–1, –2).
(
(
(
)
)
)
y
7
x
3
0
P (x, y)
E(–7, –5)
(b) Let Q(11, 4) divides EG in the ratio m : n.
The y-coordinate of Q is 4,
n(–5) + m(5)
=4
m+n
5m – 5n = 4m + 4n
m = 9n
m 9
=
n 1
G(13, 5)
CHAPTER 7
◆
y
n
m
0
G(13, 5)
Q(11, 4)
x
E(–7, –5)
Thus, the ratio EQ : QG is 9 : 1.
7.1.3
181
4 . Making a conclusion
(a) EP = !(–1 – (–7))2 + (–2 – (–5))2
PG = !(13 – (–1))2 + (5 – (–2))2
2
2
6 + 3 = !w
142 + 72
= !w
= !w
45= !w
245
5 = 7!w
5
= 3!w
5 : 7!w
5 = 3 : 7.
Thus, the ratio EP : PG = 3!w
(b) EQ = !(11 – (–7))2 + (4 – (–5))2
QG = !(13 – 11)2 + (5 – 4)2
182 + 92= !w
22 + 12
= !w
= !w
405= !w
5
5
= 9!w
5 : !w
5 = 9 : 1.
Thus, the ratio EQ : QG = 9!w
Self Practice 7.3
CHAPTER 7
1. The diagram on the right shows the position of two football
players, P and Q. The coordinates of players P and Q are
(4, 6) and (40, 45) respectively. Player P wants to kick the
2
ball to player Q but the ball lands at the distance of the
3
straight line towards player Q from player P. Determine
the coordinates of the ball when it touches the surface of
the field.
2. The diagram on the right shows the plan of a straight
highway between two towns, A and B on the Cartesian
plane. An engineer wants to build two rest houses between
the two towns such that the two rest houses divide the
road into three parts of equal distance. Determine the
coordinates of the two rest houses.
3. The diagram on the right shows the position of two
ducklings, H and K on the Cartesian plane. Given the
coordinates of duckling H are (–3, –2) and the coordinates
of duckling K are (6, 10). The two ducklings walk towards
each other with different velocities and meet at point L.
The velocity of duckling H is twice the velocity of
duckling K.
(a) State the ratio HL : LK.
(b) Find the distance of duckling K from its original
position when duckling K meets duckling H.
182
y
Q(40, 45)
0
P(4, 6)
x
y
A(–4, 5)
B(5, 2)
x
0
y
K(6, 10)
H(–3, –2) 0
x
7.1.3
Coordinate Geometry
Intensive Practice 7.1
Scan QR code or visit bit.ly/2pT22Zk for the quiz
1. A straight line passes through P(2, 8) and Q(7, 3). Point R divides line segment PQ such that
PR = 4QR. Find the coordinates of point R.
2. If the point R(6, 3) divides the line segment from P(4, 5) to Q(x, y) in the ratio 2 : 5, find
(a) the coordinates of Q,
(b) the coordinates of the midpoint of PQ.
3. Point C(1, 4) divides the straight line joining points A(–3, 6) and B(h, k) in the ratio 2 : 3.
Find the value of h and k.
4. Points A(4r, r), B(e, f) and C(3e, 4f) are on a straight line. B divides the straight line AC in the
ratio 3 : 4. Express e in terms of f.
5. The diagram on the right shows a quadrilateral PQRS
with vertices P(1, 4), Q(7, –8), R(9, 5) and S(5, 1).
Point T(6, 2) is on the straight line RS. Find
(a) the coordinates of point U which divides the side PQ
in the ratio 2 : 1,
(b) the coordinates of the midpoint of side QR,
(c) the ratio RT : TS,
(d) the length of side PS.
y
0
P(1, 4)
T(6, 2)
R(9, 5)
S(5, 1)
x
6. Point P(k, 2) divides the straight line joining the points A(−2, 1) and B(2, 5) in the ratio m : n.
Find
(a) the ratio m : n,
(b) the value of k.
7. The diagram below shows the position of the multipurpose hall, school, market and bus stop
on a Cartesian plane. Haziq’s house is at the midpoint of P1P2 such that P1 divides the line
segment from the multipurpose hall to the market in the ratio 4 : 1, while P2 divides the line
segment from the school to the bus stop in the ratio 1 : 2.
12
10
8
6
4
2
–5
y
Multipurpose
hall
0
–2
Bus stop
School
5
10
Market
x
15
Determine the position of Haziq’s house.
183
CHAPTER 7
Q(7, –8)
7.2
Parallel Lines and Perpendicular Lines
It is easy to find parallel lines and perpendicular lines around
us. The floats separating the lanes in a swimming pool and the
support structures used in construction are some of the examples
of parallel and perpendicular lines. What are some other examples
of parallel and perpendicular lines around us?
Making and verifying conjectures about gradient of parallel
and perpendicular lines
INQUIRY 2
In groups
21st Century Learning
CHAPTER 7
Aim: To make and verify conjectures about the relationship between the gradient of two
parallel lines and the gradient of two perpendicular lines
Instructions:
1. Form two groups and each group will choose one activity.
Activity 1
1. Using GeoGebra software, draw straight lines L1 and L2 that are parallel to each
other on the Cartesian plane.
2. Record the gradient of straight lines L1 and L2.
3. Drag straight lines L1 or L2 and observe the changes on the gradient of L1 and L2.
4. What can you say about the relationship between the gradient of straight lines L1
and L2?
5. Measure the angles formed between the lines L1 and L2 and the positive x-axis.
What can you observe about the two angles? Explain.
6. With your group members, verify the relationship you obtained in step 4 with
the results you obtained in step 5.
Activity 2
1. Using GeoGebra software, draw straight lines L1 and L2 that are perpendicular to
each other on the Cartesian plane.
2. Record the gradient of L1 and L2 and determine the product of gradient of L1
and L2.
3. Drag straight lines L1 or L2 and observe the change in the gradient of L1 and L2 as
well as the product of their gradient.
4. What can you say about the relationship between the gradient of L1 and L2?
5. Measure θ1 and θ2, the angles formed between lines L1 and L2 respectively
towards positive x-axis. Hence, determine the product of tan θ1 and tan θ2.
6. What is the relationship between tan θ1 and tan θ2? Explain.
7. With your group members, verify the relationship you obtained in step 4 with
the results you obtained in step 6.
2. Appoint a representative from each group to present the results obtained in front of
the class.
184
7.2.1
Coordinate Geometry
We have learnt that the gradient, m of a straight line L which passes
point A(x1, y1) and B(x2, y2) is given by the formula:
Gradient, m =
y
L
B(x2, y2)
y2 – y1
x2 – x1
As shown in the diagram on the right, in ΔABC,
y2 – y1 BC
=
x2 – x1 AC
m = tan θ
So, the definition of gradient, m of a straight line is:
0
y2 – y1
A(x1, y1) θ
C
θ x2 – x1
x
m = tan θ
with θ being the angle formed between a straight line and the positive x-axis and
0°  θ < 180°.
The following diagrams show the gradient of a straight line L changing when θ increases
from 0º to 180º.
y
L2
L3
L1
x
0
θ = 0°,
tan θ = tan 0°
mL = 0
1
0
x
θ = 90°,
tan θ = tan 90°
mL not defined
2
θ
0
y
L4
θ
x
0
x
90° < θ < 180°,
tan θ < 0
mL < 0
0° < θ < 90°,
tan θ > 0
mL > 0
4
3
Thus, from activity 1 in Inquiry 2, let m1 and m2
be the gradient of straight lines L1 and L2 respectively. If lines L1 and L2
are parallel, then
θ1 = θ2
Corresponding angles, lines //
tan θ1 = tan θ2
that is, m1 = m2
Conversely, if m1 = m2, we can find that θ1 = θ2 and L1 is parallel to L2.
y
L1
L2
θ1
0
θ2
x
Two straight lines, L1 and L2 are parallel to each other if and only if m1 = m2.
From the results of activity 2 in Inquiry 2, let m1 and
m2 be the gradient of straight lines L1 and L2 respectively and
θ1 � 0.
y
Exterior angle
θ2 = 90° + θ1
tan θ2 = tan (90° + θ1) of Δ
L1
1
tan θ2 = –
C
tan θ1
θ2
tan θ1 tan θ2 = –1
θ1
that is, m1m2 = –1
A
0 B
L2
7.2.1
Smart TIPS
x
In DABC,
tan θ1 = AC
BC
BC = 1
AC tan θ1
So, tan θ2 = – BC
AC
J
=– 1
tan θ1
185
CHAPTER 7
y
y
Conversely, if m1m2 = –1 we can find that θ2 = 90° + θ1 and L1 is perpendicular to L2.
Two straight lines, L1 and L2 are perpendicular to each other if and only if m1m2 = –1.
Example 5
(a) Show whether the straight lines 6x + 9y = 7 and
x y
+ = 1 are parallel.
3 2
k
x, such that k is a constant is parallel to the straight line
3
2x + 3y = 9. Find the value of k.
(b) The straight line y = 4 –
CHAPTER 7
Solution
(a) Write equation 6x + 9y = 7 in gradient form.
6x + 9y = 7
9y = – 6x + 7
2
7
Arrange in gradient form, y = mx + c
y =– x+
3
9
2
Gradient, m1 = –
3
x y
Straight line in intercept form
For straight line equation + = 1,
3 2
b
Gradient, m2 = –
a
2
=–
3
Since both straight lines have the same gradient, they are parallel.
k
(b) y = 4 – x
3
k
y=– x+4
3
MATHEMATICS
k
Gradient, m1 = –
POCKET
3
• Gradient form
2x + 3y = 9
y = mx + c, where m
3y = –2x + 9
is gradient and c is
2
y-intercept.
y = – x + 3
3
• Intercept form
2
x + x = 1, where a
Gradient form, y = mx + c
Gradient, m2 = –
a b
3
and b are the x-intercept
Since both of the straight lines are parallel,
and y-intercept
m1 = m2
respectively, and the
k
2
gradient is – b .
– =–
a
3
3
k=2
186
7.2.1
Coordinate Geometry
Example 6
(a) Determine whether straight lines y – 3x = 5 and 3y + x – 12 = 0 are perpendicular.
(b) The vertices of a triangle ABC are A(0, –5), B(2, 1) and C(–7, k), such that k is a constant.
Find the value of k if ∠ABC = 90°.
Solution
(a) Write both equations in the gradient form to find their gradients.
y – 3x = 5
y = 3x + 5
Gradient, m1 = 3
3y + x – 12 = 0
3y = –x + 12
1
y=– x+4
3
1
Gradient, m2 = –
3
1
It is found that, m1m2 = 3 –
3
= –1
Thus, straight lines y – 3x = 5 and 3y + x – 12 = 0 are perpendicular to each other.
( )
(1 2– –(–5)0 )(–7k ––12) = –1
k–1
= –1
3(
–9 )
k–1=3
k=4
C(–7, k)
y
0
B(2, 1)
x
Mind Challenge
Can Pythagoras Theorem be
used to verify the answer in
Example 6(b)?
CHAPTER 7
(b) Since ∠ABC = 90°,
mAB mBC = –1
A(0, –5)
Self Practice 7.4
1. Determine whether the following pairs of straight line are parallel or perpendicular to each
other.
3
(a) 2x + 3y = 9 and 4x + 6y = 0
(b) y = x – 5 and 4y – 3x = 12
4
(c) x – 2y = 6 and 2x + y = 5
(d) 2x + 3y = 9 and 2y = 3x + 10
2. The following pairs of straight lines are parallel, such that p is a constant. Find the value of p.
x y
(a) 2y = 10 – x and y = 3px – 1
(b)
– = 1 and py = 4x – 6
3 6
3. The following pairs of straight lines are perpendicular to each other. Find the value of constant k.
x y
(a) 3x + 5y = 15 and 5x – ky = 2
(b)
+ = 1 and ky = 2x – 7
3 9
4. The vertices of a triangle ABC are A(1, 1), B(–1, 4), and C(5, a). Find the value of constant a
if AB is perpendicular to BC.
7.2.1
187
Solving problems involving equations of parallel and perpendicular
lines
y
Application
im
a
C
D
el
r
te
Su
n
la
Ja
a4
ra
te
Su
x
5
Solution
A
E
n
la
Ja
The diagram on the right shows the position
of roads AD, BC and EF drawn on a Cartesian
plane. AD and BC are perpendicular to each other
and they meet at a traffic light junction, while
BC and EF are parallel to each other. Given the
coordinates of A are (18, 16) and F(20, −1) while
the equation of road BC is 5y + 4x = 70, find
(a) the equation of road EF,
(b) the equation of road AD,
(c) the coordinates of the traffic light.
an
Mathematics
Ja
l
Example 7
0
D
B
F
1 . Understanding the problem
Road AD and BC are perpendicular.
Road BC and EF are parallel.
Coordinates of point A are (18, 16), F are (20, –1) and the equation of road BC is
5y + 4x = 70.
◆ Find the equation of roads EF and AD as well as the coordinates of traffic lights
which are at the junction of roads AD and BC.
◆
◆
◆
CHAPTER 7
2 . Planning the strategy
◆
◆
◆
◆
◆
Write the equation 5y + 4x = 70 in the gradient form to determine its gradient, m1.
Use m1 = m2 to find the gradient of road EF.
Use formula m1m2 = –1 to find the gradient of road AD.
Use formula y – y1 = m(x – x1) to find the equation of roads EF and AD.
Solve equation 5y + 4x = 70 and equation AD simultaneously to find the coordinates of
the traffic lights.
3 . Implementing the strategy
(a)
5y + 4x = 70
5y = – 4x + 70
4
y = – x + 14
5
4
Gradient, m1 = – , thus gradient
5
4
EF which is parallel to BC is – .
5
188
Equation of road EF that passes
through point F(20, –1) is
4
y – (–1) = – (x – 20)
5
5y + 5 = – 4x + 80
5y + 4x = 75
7.2.2
Coordinate Geometry
4
(b) Gradient, m1 = – , thus gradient of
5
road AD, m2 which is perpendicular is
(c) Equation of BC: 5y + 4x = 70 …
Equation of AD: 4y – 5x = –26 …
25y + 20x = 350 …
1 × 5:
4
16y – 20x = –104 …
2 × 4:
– m2 = –1
5
41y = 246
3 + 4:
5
y=6
m2 =
4
Substitute y = 6 into (1).
Equation of road AD which passes through 5(6) + 4x = 70
30 + 4x = 70
point A(18, 16) is
4x = 40
5
y – 16 = (x – 18)
x = 10
4
Thus, the coordinates of traffic lights
4y – 64 = 5x – 90
are (10, 6).
4y – 5x = –26
1
2
3
4
4 . Making a conclusion
Substitute point F(20, −1) into equation
5y + 4x = 75.
Left side = 5(−1) + 4(20)
= 75
= right side
FLASHBACK
Equation of straight line
with gradient m and passing
through point (x1, y1) is
y – y1 = m(x – x1)
Thus, 5y + 4x = 75 is the equation of road EF.
CHAPTER 7
Substitute point A(18, 16) into equation
4y – 5x = –26.
Left side = 4(16) – 5(18)
= –26
= right side
Thus, 4y – 5x = –26 is the equation of road AD.
From the graph on the right, the coordinates of
traffic lights are (10, 6).
C
16
12
8
4
y
5y + 4x = 70
A
(10, 6)
–8 –4 0 4 8 12 16 20
–4 4y – 5x = –26
B
–8
D
7.2.2
x
189
Self Practice 7.5
B(8, 12)
1. In the diagram on the right, radius AB of the Ferris wheel is
perpendicular to the tangent of the circle at point B(8, 12).
The equation of the tangent to the circle at point B is given
as 3x + 2y = 48. Find the equation of radius AB of the
Ferris wheel.
2. The diagram on the right shows the plan of a rectangularshaped hut drawn on a Cartesian plane. A pipe of the
shortest length will be connected from the main pipe at
P(8, 7) to the hut. Find
(a) the coordinates of point of connection of the pipe at
the hut,
(b) the length of trench that should be dug in order to bury
the pipe to the hut.
Intensive Practice 7.2
A
y
A(3, 8)
P(8, 7)
B(7, 2)
x
0
Scan the QR code or visit bit.ly/2Vtz9P4 for the quiz
CHAPTER 7
1. For each of the following, determine whether the lines AB and CD are parallel or
perpendicular to each other.
(a) A(6, 2), B(3, 4), C(3, –1), D(–3, 3)
(b) A(4, –3), B(–3, 4), C(1, 4), D(–2, 1)
2. Given A(1, 2), B(6, 8) and C(12, k) are the vertices of a triangle, such that
∠ABC = 90°, find the value of k.
3. Given P(7, 3), Q(2, 2) and R(−1, 4). Find
(a) the equation of straight line that passes through point P and is parallel to QR,
(b) the equation of straight line that passes through point R and is perpendicular to QR.
Then, find the coordinates of S such that both lines intersect.
4. The coordinates of three points are P(−1, −6), Q(3, −12) and R(e, 6). Find the value of
constant e if
(a) P, Q and R are collinear,
(b) PQ is perpendicular to PR.
5. Given four points, P(– 6, 1), Q(1, –2), R(0, 5) and S(–3, h). If PQ is perpendicular to RS, find
the value of constant h.
y
6. In the diagram on the right, OC is perpendicular from origin
O to straight line AB, such that point A is (0, 5) and point B is
(10, 0). Find
(a) the equation of straight line AB and OC,
(b) the coordinates of C and distance OC.
190
A(0, 5)
O
C
B(10, 0)
x
7.2.2
Coordinate Geometry
3y – x = 15
D(3, 6)
A(1, 2)
C
B
E
x
0
8. The diagram on the right shows a trapezium ABCD. The
coordinates of A, B, C and D are A(3, 6), B(11, 10), C(9, 4)
and D(5, 2) respectively.
(a) Determine the pairs of parallel and perpendicular lines.
(b) Find the equation of straight line AB.
(c) A straight line passes through point C and is perpendicular
to AB. Find the equation of the straight line. Show that the
straight line passes through the midpoint of AB.
9. The diagram on the right shows a triangle ABC, such that
A(9, 9) and C(1, –3). Point B lies on the perpendicular bisector
of AC and the equation of straight line AB is y = 8x – 63.
(a) Find
(i) the equation of perpendicular bisector of AC,
(ii) the coordinates of B.
(b) Point D lies on the diagram such that ABCD is a
rhombus.
(i) Find the coordinates of D.
(ii) Show that AC = 2BD.
10. In the diagram on the right, two straight lines, y = x + 4 and
3y = x + 8 intersect at point P. The straight line that passes
through point P with gradient 2 meets the x-axis and y-axis at
points A and B respectively. Show that
(a) the coordinates of P are (–2, 2),
(b) the equation of straight line that passes through point P
and is perpendicular to straight line AB is 2y + x = 2,
(c) the coordinates of A are (–3, 0) and the coordinates of B
are (0, 6),
AP
1
is .
(d) the ratio of
PB
2
y
B(11, 10)
A(3, 6)
C(9, 4)
D(5, 2)
x
0
y
A(9, 9)
y = 8x – 63
B
0
x
C(1, –3)
y
y=x+4
B
3y = x + 8
P
A
m=2
0
11. In the diagram on the right, BC is a tangent to the circle with
centre A(– 4, –2) at point B(1, 8). Find the equation of tangent
BC.
x
B(1, 8)
C
A(–4, –2)
191
CHAPTER 7
7. The diagram on the right shows a parallelogram ABCD.
Points A and D are (1, 2) and (3, 6) respectively. Equation of
the straight line DC is 3y – x = 15. DE is the perpendicular
bisector of AB. Find
(a) the equation for AB and DE,
(b) the coordinates of E and B.
y
7.3
Areas of Polygons
Deriving formula of area of triangles
We can use formula to find the area of a polygon on the Cartesian plane if the vertices are
known. Follow the exploration below to derive the formula of area of a triangle when the
coordinates of each vertex are known.
InQuirY 3
In groups
21st Century Learning
CHAPTER 7
Aim: To determine area of triangle when coordinates of each vertex are known
Instructions:
1. Using GeoGebra software, draw a triangle with vertices A, B, and C.
2. Draw dashed lines as shown in the diagram on the right.
y
3. Using the instruction menu in the software,
(a) find the length of AD, DE, BE and CD.
C(2, 6)
6
(b) find the area of ΔACD, trapezium BCDE and ΔABE.
(c) determine the area of ΔABC by using the values
4
B(6, 3)
obtained in (b).
2
4. Discuss with your group members the way to obtain
x
area of the triangle.
6
–2 0
2 4
5. Are there any other ways to obtain the area of triangle A(–2, –2)
–2
D
E
ABC?
From the results of Inquiry 3, we can make a generalisation about the method to obtain area of
a triangle by using the following formula.
Diagram 7.2 shows a triangle ABC, with the position of A(x1, y1), B(x2, y2) and C(x3, y3)
arranged in order.
Area of ΔABC
= area of ΔACD + area of trapezium BCDE – area of ΔABE
1
1
1
× AD × CD +
× DE × (BE + CD) –
× AE × BE
=
2
2
2
1
1
= (x3 – x1)(y3 – y1) + (x2 – x3)[(y2 – y1) + (y3 – y1)]
2
2
1
– (x2 – x1)(y2 – y1)
2
1
= (x3y3 – x3y1 – x1y3 + x1y1 + x2y2 – x2y1 + x2y3 – x2y1 – x3y2
2
(
) (
) (
+ x3y1 – x3y3 + x3y1 – x2y2 + x2y1 + x1y2 – x1y1)
1
= (x1y2 + x2y3 + x3y1 – x2y1 – x3y2 – x1y3)
2
192
)
y
C (x3, y3)
B (x2, y2)
x
0
A (x1, y1)
D
E
Diagram 7.2
7.3.1
Coordinate Geometry
x1
y1
|
|
with the sum of all the products in the
direction is given
1
the positive sign, that is (x1y2 + x2y3 + x3y1) and the sum of all
2
the products in the
direction is given the negative sign, that is
1
(–x y – x y – x y ).
2 2 1 3 2 1 3
Thus, formula for area of a triangle ABC with vertices A(x1,
y1), B(x2, y2) and C(x3, y3) arranged in order can be written as:
x2
x3
x1
1 x1
2 y1
y2
y3
y1
1
= |x1y2 + x2y3 + x3y1 – x2y1 – x3y2 – x1y3|
2
1
= |(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)|
2
Area of ∆ABC =
|
|
Determining the area of triangles by using
the formula
Example 8
Find the area of DABC with vertices A(– 4, – 6), B(5, 3) and
C(2, 8).
Solution
If the coordinates are arranged anticlockwise,
y
area of ΔABC
C (2, 8)
5
2
–4
1 –4
=
2 –6
3
8
–6
B (5, 3)
1
= |(–12 + 40 – 12) – (–30 + 6 – 32)|
x
2
0
1
= |72|
2
A (– 4, – 6)
= 36 units2
If the coordinates are arranged clockwise,
2
5
–4
1 –4
area of ΔABC =
2 –6
8
3
–6
1
= |(–32 + 6 – 30) – (–12 + 40 – 12)|
2
1
Take absolute value
= |–72|
2
1
= (72)
2
= 36 units2
|
|
|
7.3.1
7.3.2
|
MATHEMATICS
POCKET
The formula on the left
is known as the shoelace
algorithm which is used
only when coordinates
of vertices are arranged
anticlockwise. If the
coordinates are arranged
clockwise, the answer
obtained would have
negative sign. In this case,
the absolute value needs to
be used so that the value
of area is positive. This
formula can be initiated by
choosing any one vertex.
Smart TIPS
Some of the quantities we
encounter in daily life have
magnitude only, such as
temperature, mass, distance,
area and volume which do
not involve direction. For
example, area of ΔABC is
36 units2. Therefore, 36 units2
is the size or magnitude
for the area of ΔABC. Such
quantities are known as
scalar quantities.
MATHEMATICS
POCKET
The shoelace formula is used
to find the area of a polygon
when the coordinates
of each vertex is known.
The two vertical lines in
this formula are known as
the absolute value which
functions to make sure that
the measurement area is
always positive. Important
note: The measurement
area of polygon accept
positive value only.
193
CHAPTER 7
This area formula can be written as:
x2
x3
1 x1
2 y1
y2
y3
Besides the shoelace algorithm, the box method can also be used to find the area of the triangle
in Example 8.
Step 1
Draw a rectangle that touches every vertex of the
triangle. Mark the triangles formed in the box with
I, II and III as shown in the diagram on the right.
y
C (2, 8)
II
III
B (5, 3)
Find the area of the rectangle by multiplying its
length by width.
x
Area of rectangle = 9 × 14
0
= 126 units2
I
Find the area of triangles I, II and III in the Step 3
rectangle.
A (– 4, – 6)
1
1
Area of triangle I = × 9 × 9 = 40 units2
2
2
1
1
Area of triangle II = × 3 × 5 = 7 units2
2
2
1
Area of triangle III = × 6 × 14 = 42 units2
BRAINSTORMING
2
Subtract each of the area of triangles obtained in
Step 4
What can you say about
step 3 from the area of rectangle to determine the
the three points A(x1, y1),
area of ΔABC.
B(x2, y2) and C(x3, y3) if the
1
1
area of
Area of ΔABC = 126 – 40 – 7 – 42
1 x1 x2 x3 x1
2
2
DABC =
= 0?
2 y y y y
= 36 units2
1 2
3 1
Step 2
|
|
CHAPTER 7
Example 9
The coordinates of the vertices of a triangle ABC are A(8, 5), B(–2, –3) and C(k, –1). If the
area of triangle ABC is 18 units2, find the possible values of k.
Solution
Since the order of the vertices of triangle ABC is not known, we need
to take the absolute value for the expression for area of triangle ABC.
1
Area of ΔABC = 8 –2 k 8
2 5 –3 –1 5
y
1
±18 = |(–24 + 2 + 5k) – (–10 – 3k – 8)|
2
1
±18 = (8k – 4)
2
1
1
(8k – 4) = –18 or (8k – 4) = 18
0
C1 (k, –1)
2
2
8k – 4 = –36
8k – 4 = 36
B (–2, –3)
8k = –32
8k = 40
k = –4
k=5
Thus, the possible values of k are – 4 and 5.
|
194
|
A (8, 5)
x
C2 (k, –1)
7.3.2
Coordinate Geometry
Self Practice 7.6
1. Find the area of the triangles with the following vertices given.
(a) (5, 10), (2, 1), (8, 3)
(b) (3, 1), (6, 4), (– 4, 2)
(c) (– 4, –3), (5, 1), (2, 6)
2. Vertices P and Q are (3, 4) and (1, –2), respectively, and vertex R is on the x-axis. Find the
coordinates of R such that area of ΔPQR is 10 units2.
3. Show that points (8, 4), (2, 1) and (–2, –1) are collinear.
4. Points E(–2, –1), F(2, p) and G(10, 5) are collinear. Find the value of p.
5. The vertices and area of ΔABC are given below. Find the possible values of k.
(a) A(− 4, −1), B(5, 3), C(−1, k); area of DABC = 15 units2
(b) A(5, k), B(3, 7), C(1, 3); area of DABC = 10 units2
(c) A(1, −2), B(k, 6), C(1, 2); area of DABC = 12 units2
(d) A(3, 0), B(4, k), C(1, 4); area of DABC = 5 units2
Determining the area of quadrilaterals by using the formula
Consider quadrilateral ABCD in the diagram on the right, with vertices
A(x1, y1), B(x2, y2), C(x3, y3) and D(x4, y4) arranged in order.
|
|
|
|
y
C (x3, y3)
|
0
A (x1, y1)
x
B (x2, y2)
CHAPTER 7
Area of quadrilateral ABCD
= area of DABC + area of DACD
1 x x x x
1 x x x x
= y1 y2 y3 y1 + y1 y3 y4 y1
2 1 2 3 1
2 1 3 4 1
1
= |(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)|
2
1
= + |(x1y3 + x3y4 + x4y1) – (x3y1 + x4y3 + x1y4)|
2
1
= |(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)|
2
1 x x x x x
= y1 y2 y3 y4 y1
2 1 2 3 4 1
D (x4, y4)
|
From the expansion of the above expression, we found that the formula obtained is similar
to the formula for area of triangle.
In general, the area of quadrilateral ABCD with vertices A(x1, y1), B(x2, y2), C(x3, y3) and
D(x4, y4) arranged in order can be written as:
x2
x3
x4
x1
1 x1
2 y1
y2
y3
y4
y1
1
= |(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)|
2
Area of quadrilateral ABCD =
7.3.2
7.3.3
|
|
195
Example 10
Find the area of quadrilateral PQRS with vertices P(3, 5), Q(–1, 3), R(2, –3) and S(4, 1).
y
Solution
Arrange the vertices in order:
Q (–1, 3)
1 3 –1 2 4 3
2 5 3 –3 1 5
1
= |(9 + 3 + 2 + 20) – (–5 + 6 – 12 + 3)|
2
1
= |42|
2
= 21 units2
Area of quadrilateral PQRS =
|
|
P (3, 5)
S (4, 1)
x
0
R (2, –3)
Self Practice 7.7
1. Find the area of the quadrilaterals with the following vertices given.
(a) (1, 7), (–5, 6), (–2, – 4) and (2, –3)
(b) (2, 9), (– 6, 4), (–1, –3) and (8, 1)
(c) (0, 2), (– 6, –2), (–3, –5) and (–1, –3)
(d) (3, 4), (–2, 0), (2, – 4) and (5, 1)
2. The vertices of a quadrilateral ABCD arranged in order are A(k, 6), B(–2, 1), C(4, 5) and
D(2, 8). If area of the quadrilateral ABCD is 30 unit2, find the value of k.
CHAPTER 7
Making generalisation about the formula of area of polygons
The idea in finding the area of triangle can be used to prove that the area of a polygon with n
sides and vertices A(x1, y1), B(x2, y2), C(x3, y3), D(x4, y4), …, N(xn, yn) is as follows.
Area of polygon
x2
x3
1 x1
=
2 y1
y2
y3
|
x4 … xn
y4 … yn
x1
y1
|
1
= |(sum of all
products) – (sum of all
products)|
2
with vertices A, B, C, D, …, N arranged in order.
In general, if the vertices A(x1, y1), B(x2, y2), C(x3, y3),
D(x4, y4), …, N(xn, yn) of a polygon with n sides are arranged
in order, then :
196
Area of polygon =
1 x1 x2 … xn x1
2 y1 y2 … yn y1
|
|
BRAINSTORMING
Surf the Internet to explore
convex polygon and
concave polygon. Discuss
with your friends whether
the formula for area of
polygon can be used for
concave polygon.
7.3.2
7.3.4
Coordinate Geometry
Example 11
Find the area of a pentagon with vertices (5, –2), (1, –1), (–1, 1), (1, 3) and (4, 4).
Solution
By plotting vertices of the pentagon as in the diagram on
the right, the vertices arranged in order would be
(4, 4), (1, 3), (–1, 1), (1, –1) and (5, –2).
y
(1, 3)
Area of pentagon = 1 4 1 –1 1 5 4
2 4 3 1 –1 –2 4
= 1 |(12 + 1 + 1 – 2 + 20) – (4 – 3 + 1 – 5 – 8)|
2
= 1 |43|
2
= 21 1 units2
2
|
|
(4, 4)
(–1, 1)
x
0
(1, –1)
(5, –2)
Self Practice 7.8
1. A pentagon ABCDE has vertices A(–2, –5), B(3, 2), C(2, 8), D(0, 9) and E(–3, 1). Find the
area of pentagon ABCDE.
2. The vertices of a hexagon are (0, –1), (–3, –1), (– 4, 2), (–2, 6), (1, 5) and (2, 1). Find the area
of the hexagon.
CHAPTER 7
Solving problems involving areas of polygons
Example 12
The vertices of a triangle ABC are A(– 4, –3), B(7, 2) and C(–2, 3). M and N
are the midpoints of the sides AC and BC respectively. Find
(a) the coordinates of M and N,
(b) the ratio of area of triangle CMN to area of quadrilateral ABNM.
Solution
(a) Coordinates of M =
= (–3, 0)
(–22+ 7, 3 +2 2 )
5 5
=( , )
2 2
Coordinates of N =
7.3.4
(– 4 +2(–2), –32+ 3)
7.3.5
y
C (–2, 3)
M
N
0
B (7, 2)
x
A (– 4, –3)
197
5
–2
2
5
3
3
0
2
= 1 0 – 15 + 15 – –9 + 0 – 5
2
2
2
1
= |14|
2
= 7 units2
5
–4 7
–3 – 4
2
1
Area of quadrilateral ABNM =
2
5
0 –3
–3 2
2
= 1 –8 + 35 + 0 + 9 – –21 + 5 – 15 + 0
2
2
2
1
= |42|
2
= 21 units2
(b) Area of triangle CMN = 1
2
|
–2
–3
|(
|
) (
|
|(
)|
|
) (
)|
Thus, the ratio of area of triangle CMN to quadrilateral ABNM is 7 : 21 = 1 : 3.
Self Practice 7.9
CHAPTER 7
1. In the diagram on the right, points A(–3, 0), B(5, 2), C and
D(–1, 6) are the vertices of a parallelogram ABCD.
M is the intersection point of diagonals AC and BD. Find
(a) the coordinates of C and M,
(b) the ratio of the area of ΔABM to the area of
parallelogram ABCD.
y
C
D (–1, 6)
M
B (5, 2)
A (–3, 0) 0
2. The straight line y = 8 – 2x intersects the straight line y = k,
x-axis and y-axis at points P, Q and R respectively. Given the
area of ΔOPR is 12 units2, with O as the origin, find
(a) the smallest value of k,
(b) the coordinates of P.
3. In the diagram on the right, ABCD is a parallelogram with
vertices A(4, –2), B(7, 5), C(4, 7) and D(k, 0). Find
(a) the area of ΔABC,
(b) the value of k if the area of ΔACD is equal to the area of
ΔABC,
(c) the coordinates of E if ACBE is a parallelogram,
(d) the area of parallelogram ACBE.
198
y
x
C (4, 7)
B (7, 5)
0
D (k, 0)
x
A (4, –2)
7.3.5
Coordinate Geometry
Intensive Practice 7.3
Scan the QR code or visit bit.ly/33hpduP for the quiz
1. ABCD is a parallelogram which diagonals intersect at E. Given A(–5, 3), B(0, –2) and
C(3, 5), find
(a) the coordinates of D and E,
(b) the area of parallelogram ABCD.
2. PQRS is a rhombus with coordinates P(3, 3), Q(h, 3), R(–5, –1) and S(0, k). Find
(a) the value of h and k,
(b) the area of rhombus PQRS.
3. Given three points, A(–1, –5), B(2, 1) and C(6, 9),
(a) find the area of ΔABC,
(b) based on the answer in (a), what can you say about points A, B and C?
4. Find the area of a polygon with vertices (5, 2), (–1, –3), (2, 6), (3, –2), (–4, 0) and (–3, 2).
5. Points A(5, –1), B(3, 3) and C(– 6, p) are the vertices of a triangle. Find the values of p if the
area of ΔABC is 16 units2.
6. Given three points, P(2, 2r – 1), Q(r – 1, r + 1) and R(r + 3, 0). If points P, Q and R are on the
same straight line, find the possible values of r.
7. Three points have coordinates A(8, a), B(–1, 2) and C(3, 10). Find the value of a if
(a) A, B and C are collinear,
(b) the area of ΔABC is 12 units2.
G (7, 8)
F (4, 4)
x
0
9. Points O(0, 0), P(m + 1, m – 7), Q(2m + 1, 2m) and R(m, m + 6) are the vertices of a
quadrilateral such that m > 0.
(a) If the area of OPQR is 34 1 units2, find the value of m.
2
(b) Hence, find the area of ΔOPR.
10. The coordinates of three LRT stations, Paya Redan,
Kampung Raja and Panjang Sari are represented by
points A(0, 9), B(7, 0) and C(12, 12), such that
1 unit represents 100 m. Find
(a) the distance, in km, between Paya Redan Station
and Kampung Raja Station.
(b) the actual area, in km2, of the triangle formed by
the three stations.
y
Paya Redan
A (0, 9)
0
Panjang Sari
C (12, 12)
B (7, 0)
Kampung Raja
x
199
CHAPTER 7
y
8. The diagram on the right shows an isosceles triangle
EFG with vertices E(0, k), F(4, 4) and G(7, 8). EF
and FG have the same length.
E (0, k)
(a) Find the value of k.
(b) H is a point on the line y = 11 such that EH = GH.
Find
(i) the coordinates of H,
(ii) the ratio of area of ΔEFG to area of quadrilateral EFGH.
7.4
Equations of Loci
The locus of a moving point is the path taken by the point subject
to certain conditions. For example, the path traced by point A
moving r units from the fixed point O on the radar screen at an
air traffic control centre as shown in the diagram on the right is
a locus in the shape of a circle and it can be represented by an
equation. Can you determine the equation of the locus of moving
point A which is a circle?
A
r
O
Determining the equation of locus
The equation of the locus of a moving point which satisfies certain conditions can be
determined using the formula for distance between two points or other formulas depending on
the given conditions.
A Locus of a moving point from a fixed point is constant
CHAPTER 7
InQuirY 4
In groups
Aim: To explore the shape and determine the equation of locus of a
moving point which the distance from a fixed point is constant
Instructions:
1. Scan the QR code or visit the link on the right.
bit.ly/2pMIhTe
2. Let P(x, y) is a point of distance r from a fixed point A(x1, y1)
where r . 0.
3. Move point P and observe the path traced by point P.
4. What is the shape of the locus of point P obtained?
5. Using the formula of distance between two points, write the equation for the shape
obtained in terms of x, y, x1, y1 and r.
From the results of Inquiry 4, the shape of locus of point P obtained is a circle with centre
A(x1, y1) and radius r unit. Equation of the locus of moving point P(x, y) which the distance
from a fixed point A(x1, y1) is always constant, can be determined using the distance formula as
follows:
Locus of P
PA = r
!(x – x1)2 + (y – y1)2 = r
(x – x1)2 + (y – y1)2 = r2, where r > 0
200
A(x1, y1)
r
P(x, y)
7.4.1
Coordinate Geometry
Example 13
Find the equation of locus of moving point P so that its distance from point A(4, –3) is 6 units.
y
Solution
5
Let the coordinates of point P are (x, y).
Distance of P from A = 6
Square both
!(x – 4)2 + [y – (–3)]2 = 6
sides of
(x – 4)2 + (y + 3)2 = 36
equation
x2 – 8x + 16 + y2 + 6y + 9 = 36
2
2
x + y – 8x + 6y – 11 = 0
Thus, equation of locus of P is x2 + y2 – 8x + 6y – 11 = 0.
0
–5
–5
x
10
A (4, –3) P (x, y)
5
– 10
B Ratio of distance of moving point from two fixed points is constant
In groups
Aim: To explore the shape and determine the equation of locus
of a moving point whose the ratio of the distance from two
fixed points is constant
Instructions:
bit.ly/2pMIjuk
1. Scan the QR code or visit the link on the right.
2. Let P(x, y) is a moving point such that its distance from two fixed points
A(x1, y1) and B(x2, y2) is in the ratio m : n, that is PA = m .
PB n
3. Drag the point on the slider to the left and right so that ratio r changes and observe the
circle formed.
4. Does the locus of moving point P form a circle? If yes, can you determine its equation in
terms of x, y, x1, y1, x2, y2, m and n?
5. Then, drag the slider r to the left again so that its value is 1, that is, PA : PB = 1 : 1.
6. Make a conjecture about the shape of locus of moving point P that will be obtained if
PA = PB. Can you determine its equation?
From the results of Inquiry 5, the shape of locus of moving point P is a circle and the equation
of locus of moving point P(x, y), which the distance from two fixed points A(x1, y1) and B(x2, y2),
is always constant in the ratio m : n can be determined by using the distance formula as follows:
PA = m
PB n
!(x – x1)2 + (y – y1)2 m
=
!(x – x )2 + (y – y )2 n
2
2
(x – x1)2 + (y – y1)2 m2
=
(x – x2)2 + (y – y2)2 n2
7.4.1
Locus of P
P(x, y)
m
n
B(x2, y2)
A(x1, y1)
201
CHAPTER 7
InQuirY 5
When PA : PB = 1 : 1, P(x, y) is always the same distance from two fixed points A(x1, y1) and
B(x2, y2), the shape of locus of P is the perpendicular bisector of line AB. Its equation is:
PA = PB
!(x – x1) + (y – y1)2 = !(x – x2)2 + (y – y2)2
(x – x1)2 + (y – y1)2 = (x – x2)2 + (y – y2)2
2
Example 14
Point P moves such that its distance from points S(1, 2) and T(4, –1) is in the ratio 2 : 1. Find
the equation of locus of moving point P.
Solution
Let P(x, y) is a moving point.
PS = 2
PT 1
!(x – 1)2 + (y – 2)2 2
=
!(x – 4)2 + (y + 1)2 1
(x – 1)2 + (y – 2)2 4
=
(x – 4)2 + (y + 1)2 1
y
2
Square both
sides of
equation
S (1, 2)
0
P (x, y)
x
5
T (4, –1)
(5, –2)
–2
–4
x2 – 2x + 1 + y2 – 4y + 4 = 4(x2 – 8x + 16 + y2 + 2y + 1)
x2 + y2 – 2x – 4y + 5 = 4x2 + 4y2 – 32x + 8y + 68
3x2 + 3y2 – 30x + 12y + 63 = 0
Divide each term by 3
x2 + y2 – 10x + 4y + 21 = 0
CHAPTER 7
Thus, the equation of locus of moving point P is x2 + y2 – 10x + 4y + 21 = 0.
Example 15
Find the equation of locus of a moving point Q such that its distance from point A(2, 3) and
point B(6, 9) are the same.
Solution
Let Q(x, y) is a moving point.
QA = QB
2
!(x – 2) + (y – 3)2 = !(x – 6)2 + (y – 9)2
(x – 2)2 + (y – 3)2 = (x – 6)2 + (y – 9)2
2
x – 4x + 4 + y2 – 6y + 9 = x2 – 12x + 36 + y2 – 18y + 81
8x + 12y – 104 = 0
2x + 3y – 26 = 0
Thus, the equation of locus of moving point Q is
2x + 3y – 26 = 0.
202
y
10
B (6, 9)
2x + 3y – 26 = 0
5
Q (x, y)
A (2, 3)
0
5
10
x
7.4.1
Coordinate Geometry
Self Practice 7.10
1. Find the equation of locus of moving point P such that its distance from each of the following
points is 3 units.
(a) (0, 0)
(b) (2, 3)
(c) (– 4, 5)
(d) (–1, – 6)
2. Point P moves such that its distance from Q(–2, 1) is always 5 units. Find the equation of
locus of point P.
3. Find the equation of locus of moving point P such that its distance from the following fixed
points is in the ratio given.
(a) A(–2, 0), B(4, 0); ratio 1 : 2
(b) C(–3, 0), D(2, 5); ratio 1 : 3
(c) E(0, 2), F(–2, 4); ratio 3 : 2
(d) R(1, 2), S(4, –1); ratio 2 : 1
4. The coordinates of points J and K are (–1, 3) and (4, 6) respectively. Point Q moves such that
QJ : QK = 2 : 3. Find the equation of locus of Q.
5. Point R moves such that its distance from point A(6, 0) is twice its distance from point
B(–3, 0). Find the equation of locus of R.
6. Point P moves in the ratio PO : PA = 1 : 4, with O the origin and coordinates of point A being
(2, 0). Find the equation of locus of point P.
7. Find the equation of locus of moving point P such that its distances from the following points
are the same.
(a) A(–2, 0) and B(0, 4)
(b) C(–3, 5) and D(2, – 4)
(c) J(2, 3) and K(6, 8)
Solving problems involving equations of loci
Mathematics
Application
A goat is tied with a rope to a pole which is planted in the
middle of a field. The length of rope used is 4 metres. The goat
walks around the pole tied to the end of the taut rope as shown
in the diagram. If the coordinates of the pole are A(2, 4), find the
equation of locus of the goat’s track.
4m
A (2, 4)
CHAPTER 7
Example 16
Solution
1 . Understanding the problem
◆
◆
◆
A goat is tied with a rope 4 metres long to the pole.
Coordinates of the pole are A(2, 4).
Find the equation of locus of goat’s track around the pole with the rope taut.
2 . Planning the strategy
Track of the goat is a circle with centre A(2, 4) and radius 4 metres.
◆ Use formula for distance between two points, d = !(x – x1)2 + (y – y1)2 to find the
equation of locus of goat’s track.
◆
7.4.1
7.4.2
203
3 . Implementing the strategy
Let P(x, y) is a moving point at the end of
rope tied to the goat’s neck.
PA = 4
!(x – 2)2 + (y – 4)2 = 4
(x – 2)2 + (y – 4)2 = 16
2
x – 4x + 4 + y2 – 8y + 16 = 16
x2 + y2 – 4x – 8y + 4 = 0
Thus, the equation of locus of moving
point P, which is the goat’s track around
the pole with the rope taut is
x2 + y2 – 4x – 8y + 4 = 0.
4 . Making a conclusion
Represent the equation of locus of P on
the Cartesian plane. It is found that the
locus of P touches the x-axis at point (2, 0).
Substitute (2, 0) into equation of locus of P.
Left side = 22 + 02 – 4(2) – 8(0) + 4
=0
= right side
y
10
8
6
4
P (x, y)
A (2, 4)
x2 + y 2 – 4 x – 8y + 4 = 0
2
0
–2
2
4
6
Self Practice 7.11
CHAPTER 7
1. The diagram on the right shows the locus of a moving
point P(x, y) that touches the x-axis at a point and is of
fixed distance from a point A(3, 4). Find the equation of
locus of P.
8 10
x
y
Locus of P
A (3, 4)
P (x, y)
x
0
2. Point P moves such that it is always the same distance from the points Q(8, 7) and R(11, 4).
Point S moves such that its distance from point T(7, 8) is always 5 units. The locus of point P
and the locus of point S intersect at two points.
(a) Find the equation of locus of point P.
(b) Show that the locus of point S is x2 + y2 – 14x – 16y + 88 = 0.
(c) Find the coordinates of the intersection points of the two loci.
3. In the diagram on the right, point A(–2, 0) and point B(1, 0)
are two fixed points. Point P moves along the circle such that
the ratio PA : PB = 2 : 1. Show that
(a) the equation of the circle is x2 + y2 – 4x = 0,
(b) point C(2, 2) is on the circle.
204
y
P (x, y)
A (–2, 0) 0
x
B (1, 0)
7.4.2
Coordinate Geometry
Intensive Practice 7.4
Scan the QR code or visit bit.ly/33hpOg3 for the quiz
1. A point R(x, y) moves so that its distance from two fixed points A(–1, 10) and B(2, 6) is such
that RA = 1 . Find
RB 2
(a) the equation of locus of R,
(b) the coordinates of the point on locus R that touches the y-axis.
y
2. The diagram on the right shows a line segment AB with
coordinates A(1, 1) and B(7, 9). Find the equation of locus
of moving point S such that triangle ABS always has a right
angle at S.
B (7, 9)
A (1, 1)
x
0
3. Point Q moves along the arc of a circle with centre (6, 5). The arc of the circle passes through
R(2, 8) and S(k, 2). Find
(a) the equation of locus of Q,
(b) the values of k.
y
Locus of P
4. The diagram on the right shows the locus of moving point P
such that its distance from point R(1, 0) and line x = –1 are the
same. Find the equation of locus of moving point P.
P (x, y)
S
x
0 R (1, 0)
5. The diagram on the right shows the x-axis and y-axis which
represent the floor and wall respectively. A pole, LK of length
9 m leaning against the wall touches the floor and wall at
points K(α, 0) and L(0, β) respectively.
(a) Write the equation which relates α and β.
(b) Given P(x, y) is a point on the pole such that the ratio
LP : PK = 1 : 2. Both ends of the pole slide along the
x-axis and y-axis. Find the equation of locus of point P.
6. A monkey is tied to one vertex of his cage which measures
1 m × 1 m with a rope. The length of the rope is 2 1 m.
2
Sketch and explain the locus if the monkey moves
anticlockwise around the cage with the rope taut.
CHAPTER 7
x = –1
y
L (0, β)
P (x, y)
Wall
0
K (α, 0)
x
Floor
2 1– m
2
1m
1m
205
Summary OF Chapter 7
Divisor of line segment
nx1 + mx2 ny1 + my2
–––––––
P(x, y) = –––––––
m+n , m+n
(
Parallel lines and
perpendicular lines
)
Lines y = m1x + c1 and y = m2x + c2
are parallel if and only if m1= m2
COORDINATE GEOMETRY
Equation of locus
• Locus of moving point P(x, y) which
distance is constant from fixed point
A(x1, y1) is (x – x1)2 + (y – y1)2 = r2
CHAPTER 7
• Locus of moving point P(x, y) which
distance from two fixed points A(x1, y1)
and B(x2, y2) is always constant in the
ratio m : n is
(x – x1)2 + (y – y1)2 m2
–––––––––––––– = –––
(x – x2)2 + (y – y2)2 n2
Lines y = m1x + c1 and y = m2x + c2 are
perpendicular if and only if m1m2 = –1
Area of polygon
1 x x x ... x x
= –– y1 y2 y3 ... yn y1
2 1 2 3
n
1
1
= –– x1 y2 + x2 y3 +, ..., + xn y1 – x2 y1 –
2
x3 y2 –, ..., – x1 yn
• Locus of moving point P(x, y) which
distance from two fixed points
A(x1, y1) and B(x2, y2) is the same is
(x – x1)2 + (y – y1)2 = (x – x2)2 + (y – y2)2
WRITE YOUR JOURNAL
Coordinate geometry has introduced the general form, gradient form, intercept form and other
forms in expressing the equation of a straight line. What are the advantages of expressing the
equation in these forms? Which form do you prefer to use? Why?
206
Coordinate Geometry
MASTERY PRACTICE
1. The diagram on the right shows a straight line AB. The
midpoint of the line joining A(h, 1) and B(7, k) is M(2, 3).
Find PL3
(a) the value of h and k,
(b) the gradient of the line,
(c) the equation of the perpendicular bisector of AB.
2. Given a straight line AB with points A(–2, 6) and B(8, – 4).
Point P lies on AB such that AP : PB = 2 : 3.
Find PL3
(a) the coordinates of point P,
(b) the equation of the straight line that is perpendicular to AB
and passes through point P.
y
B (7, k)
M (2, 3)
A (h, 1)
x
0
y
A (–2, 6)
P
x
0
B (8, – 4)
3. Given three points P(1, –1), Q(n, 2) and R(n2, n + 3). If points P, Q and R lie on the same
straight line, find the possible values of n. PL3
4. Given two points R(–3, 4) and S(3, –1). Find the coordinates of point T that might be on the
y-axis such that area of ΔRST is 13.5 unit2. PL3
y
6. The diagram on the right shows the position of two birds,
A and B on the Cartesian plane. The coordinates of birds A
and B are (–3, –1) and (6, 5) respectively. The two birds fly
towards each other in a straight line with different velocities.
The velocity of bird A is twice the velocity of bird B. Find the
coordinates where the two birds meet. PL3
0
Bird B
x
Bird A
y
7. The diagram on the right shows an isosceles triangle ABC
such that the coordinates of A are (2, 2), coordinates of B are
D
(6, 2) and C is below the x-axis. PL3
(a) Given the area of ΔABC is 10 unit2, find the coordinates of C.
A (2, 2)
(b) Line CB is extended to point D so that B is the midpoint
B (6, 2)
of CD. Find the coordinates of D.
x
0
(c) A line is drawn from point D, parallel to AC, to point
E (11, k)
E(11, k) and C is joined to E.
C
(i) Find the value of k.
(ii) Show that CED is not a right-angled triangle.
207
CHAPTER 7
5. Point P(x, y) moves such that its distance from point A(2, 0) is three times its distance from
point B(– 4, 0). Find the equation of locus of point P. PL3
8. In the diagram on the right, PQRS is a trapezium with
PQ parallel to SR and ∠PQR = 90°. The coordinates
of vertex R are (11, 7). The equation of PQ and PS are
1
y = x and y = 2x – 5 respectively. Find PL4
3
(a) the coordinates of P,
(b) the equation of QR and SR,
(c) the coordinates of Q and S,
(d) the area of trapezium PQRS.
area ΔPQR PQ
=
.
Hence, show that
area ΔPRS SR
y
S
Q
P
0
9. The coordinates of the vertices of ΔJKL are J(2, 1),
K(11, 5) and L(5, 9). Point P(h, k) is in the triangle
such that all the areas of ΔJKP, ΔKLP and ΔJLP
are the same. PL5
(a) Find the area of ΔJKL.
(b) Express the area of ΔJKP and ΔKLP in terms of h and k.
(c) Find the coordinates of P.
(d) Find the equation of the line JP.
(e) If JP is extended to meet KL at Q, find
(i) the coordinates of Q,
(ii) the ratio of KQ : QL.
10. In the diagram on the right, POQ is the graph of y = ⎟2x⎢.
R is a point on OP such that OR = !w
45 unit and O is the
origin. RS is perpendicular to OP and OR is parallel to TS.
Find PL5
(a) the coordinates of R, S and T,
(b) the area of trapezium ORST.
CHAPTER 7
R (11, 7)
y = 2x – 5
y
x
L (5, 9)
P (h, k)
0
1 x
y=—
3
K (11, 5)
J (2, 1)
x
y
S
P
Q
R
y = |2x|
T
O
11. P(h, 8) and Q(k, 2) are two points on the curve y =
8
.
x
x
PL5
(a) Find the value of h and k.
(b) Find the equation of PQ.
(c) Using coordinate geometry methods, find the equations of tangent to the curve that are
parallel to PQ.
12. The diagram on the right shows a parallelogram ABCD
with the coordinates of A and B being (−8, 5) and (−2, 1)
respectively. The equation of AD is 5y – x = 33. Line PB is
perpendicular to line AD and AP : PD = 1 : 2. Find PL4
(a) the equation of BP,
(b) the coordinates of P, D and C,
(c) the area of parallelogram ABCD.
208
y
A (–8, 5) P
B (–2, 1) 0
5y – x = 33
D
C
x
Coordinate Geometry
13. In the diagram on the right, AC and BD are the diameter of
the circle with centre E. Points A, C and D are (−1, −2), (7, 4)
and (0, 5) respectively. PL5
(a) Find the coordinates of E and B.
(b) What type of quadrilateral is ABCD?
y
D(0, 5)
C(7, 4)
E
x
0
A(–1, –2)
B
14. Early of each month, a magazine publishing company sells x copies of magazines at RM6.00
per copy. The cost for one copy of magazine is RM2.00 and early every month, the company
pays a fixed cost of RM400 for printing, storage and delivery. PL5
(a) Write the equation which relates profit, P, in RM, to the number of copies, x of magazines
sold.
(b) Draw a graph for the equation obtained. From the graph drawn,
(i) find the profit gained if 500 copies of magazines are sold,
(ii) calculate the number of copies of magazines sold if the profit gained is RM1 000.
15. The vertices of a triangle ABC are A(1, 2), B(6, 7) and C(7, 2). Draw triangle ABC and
construct the perpendicular bisector of AB, BC and CA. Mark the intersection point as P.
What can you say about the intersection point? Draw a circle with centre P and radius AP.
What can you say about the circle? Repeat the same procedures for other triangles in order to
verify your answer. PL6
MATHEMATICS
1. The equation y = mx where m is the gradient, defines a group of lines, that is one line for
every value of m.
(a) Using dynamic geometry software, draw the graphs for the group of lines when its
1
gradient is zero, m = 0, followed by positive gradient, that is m = , m = 1,
2
1
m = 2 and m = 6, then negative gradient, that is m = – , m = –1, m = –2 and
2
m = – 6.
(b) From the graphs obtained, what happens to the magnitude of gradient of the lines
when the graphs get nearer to the vertical line? Can you draw a conclusion about each
line in the group of lines?
2. The equation y = 2x + c defines a group of lines, that is a line for every value of c.
(a) Using dynamic geometry software, draw the graph for the group of lines when c = – 6,
c = –3, c = 0 and c = 6.
(b) From the graph drawn, what can you conclude about every line of the group of lines?
209
CHAPTER 7
Exploring
CHAPTER
Vectors
8
What will be learnt?
Vectors
Addition and Subtraction of Vectors
Vectors in Cartesian Plane
List of
Learning
Standards
bit.ly/33rCukN
CHAPTER 8
wORDS
KEY W
Vector
Magnitude
Direction
Directed line
segment
Zero vector
Negative vector
Collinear
Resultant vector
Position vector
Triangle law
Parallelogram law
Polygon law
210
Vektor
Magnitud
Arah
Tembereng garis
berarah
Vektor sifar
Vektor negatif
Segaris
Vektor paduan
Vektor kedudukan
Hukum segi tiga
Hukum segi empat selari
Hukum poligon
The flight system in Malaysia connects people
to different destinations around the world. The
company extended flight routes to more than 1 000
destinations, involving around 150 countries. In your
opinion, what kind of information is needed for a
pilot to make sure that the suitable route is selected
for the intended destination?
8.1.1
Vectors
Did you Know?
For a quantity that involves magnitude and
direction, vector is applied widely in the fields
of mathematics and physics. Other than that,
vector is also applied in daily life such as in
navigations, computer science, geometry and
topology fields.
For further information:
bit.ly/2MCJKUa
SIGNIFICANCE
OF THIS CH
APTER
CHAPTER 8
Knowledge of vector is important because
of its application in the field of mathematics
and physics. In the branch of mechanics,
vector is used to represent quantity such
as displacement, force, weight, velocity and
momentum. Vector is also widely used in sailing
and flight.
Scan this QR code to
watch a video about
Malaysia Airlines.
bit.ly/2LazgOc
8.1.1
211
8.1
Vectors
Comparing the differences and identifying vector and scalar
In our daily lives, there are various quantities with magnitude and
direction and quantities with magnitude but without direction.
A quantity that has magnitude and direction is called a vector
quantity, while a quantity with magnitude but no direction is
called a scalar quantity.
QR
Additional information
about vector and scalar.
Observe the following situations carefully.
Temperature of a liquid inside a
refrigerator is –12°C.
bit.ly/2OCQiVm
A moving car on the road is heading
south with speed of 80 km h−1.
Mind Challenge
Can you determine which situation involves a vector quantity and
a scalar quantity? How do you identify whether the quantity is a
vector quantity or a scalar quantity?
List a few situations that
involve vector and scalar and
list a few situations that do
not involve vector or scalar.
CHAPTER 8
The following table shows the examples of quantities with
vector and scalar and quantities without both vector and scalar.
Vector
Scalar
Without vector
and scalar
50 N of force
applied to a box.
Auni's height is
1.48 m.
Pressure and
tension
Velocity of a car is
90 km h–1 and is
heading east.
The area of a tile is
120 cm2.
Conductivity of
metal
212
BRAINSTORMING
Scalar quantity is tensor
at level zero while scalar
vector is tensor at level
one. Surf the Internet to
look for further information
regarding to tensor and
discuss your findings.
8.1.1
Vectors
Can you differentiate between distance and displacement, speed and velocity and also mass and
weight? The following table shows the differences between those quantities.
Scalar quantity
Vector quantity
Examples
Distance
Total length of track that is
passed by a moving object.
Displacement
The length of the shortest
line segment between
the starting point and the
terminal point and involves
the direction from a
reference point.
30 km
40 km
Finish
50 km
Start
A car moves 40 km to the North
and 30 km to the East.
Distance = 40 km + 30 km
= 70 km
Displacement = 50 km
Mass
Amount of matter that an
object contains. The value
does not change according
to the location.
8.1.1
Velocity
Rate of change of
displacement against
time. The value becomes
negative if the object is
moving in the opposite
direction.
Weight
The pull of gravity on an
object. The value is not
constant and depends on
the gravitational force of
the location.
A
B
Haziq is moving from A to B
with the same speed and velocity,
which is 90 km h–1. After that he
turns back from B to A with speed
of 90 km h–1 and velocity of
–90 km h–1.
The mass of an astronaut when
he is on the moon is 120 kg with
weight of 200 N while the mass
of the astronaut is 120 kg and
weight is 1200 N when he is on
earth.
213
CHAPTER 8
Speed
Rate of change of distance
against time.
Example 1
State whether each of the following quantities is a vector quantity or a scalar quantity. Justify
your answer.
(a) Mikail walks 1 km from home to the grocery store.
(b) A car is moving at a speed of 90 km h–1 to the south.
(c) The body temperature of Alicia hits 38°C.
Solution
(a) Scalar quantity because the quantity only consists of magnitude.
(b) Vector quantity because the quantity consists of magnitude and direction.
(c) Scalar quantity because the quantity only consists of magnitude.
Self Practice 8.1
1. State whether each of the following quantities is a scalar quantity or a vector quantity. Justify
your answer.
(a) Liquid X with a density of 1.2 g cm−3.
(b) A box weight 150 N is moved as high as 1 m from the floor.
(c) The volume of a mineral water bottle is 1.5 l.
(d) The duration of holiday for Suzie is 3 days and 2 nights.
(e) A ball is given a horizontal impulse of 5.0 Ns.
CHAPTER 8
Representing the vector using directed line segment and vector
notations and determining the magnitude and direction of vectors
Vector can be represented using a line segment with an arrow or
better known as directional line segment. The arrow represents the
direction of the vector while the length of the line represents the
magnitude of the vector.
As an example, vector of a sailing boat moves 7 km to the east
from point A to point B can be represented by the directional line
segment as shown in the diagram on the right. Point A is the starting
point and Point B is the terminal point.
Vector can also be represented with the notation as below:
→
A
AB or AB
a or a
~
B
→
AB or AB or ∼
a or a
Magnitude of the vector can be written as:
→
|AB| or |AB| or a
∼ or |a|
214
8.1.1
8.1.2
Vectors
Zero vectors are vectors that consist of zero magnitude and the
direction cannot be determined. Zero vectors can be represented
as ∼
0.
Example:
A race car is moving in the round-shaped track. The starting point
and the terminal point of the moving car are the same. Thus, the
vector for displacement of the race car is zero vectors.
Two vectors are the same if both vectors consist of same
→ →
magnitude and direction, AB = CD.
Example:
Rakesh and Fauzi are riding bicycles at the same speed and in the
same direction. The velocity vector, v, for both movements are
the same. Thus, vRakesh = vFauzi.
A vector is negative if the vector consists of the same magnitude
→
but is moving in the opposite direction. Vector BA is a negative
→
→
→
vector of vector AB and is written as BA = – AB.
Example:
Two trains, A and B move pass each other at two parallel
railways at the same speed but different direction. The velocity
of train A has a positive value while the velocity of train B has a
negative value.
CHAPTER 8
Example 2
Draw and label the following vectors.
→
(a) SR represents displacement of 12 km to the east.
(b) p represents force of 7 N to the south.
∼
(c) r represents velocity of 70 m s−1 to the left.
Solution
(a)(b)(c)
N
N
S
p
~
R
1 cm represents 6 km
1 cm represents 3.5 N
8.1.2
r
1 cm represents 20 m s–1
215
Example 3
→
The diagram on the right shows vector AB that represents the
displacement of a particle from point A to point B. Find the
magnitude and direction of the particle from point A.
B
N
Solution
A
1 cm
1 cm
→
AB = !w
52 + 52
= !w
50
2
= 5!w
→
→
2 cm and the direction of AB is to the Northeast.
Thus, the magnitude of AB is 5!w
Example 4
The diagram on the right shows a parallelogram, ABCD. The
points P, Q, R and S are the midpoints of AB, BC, CD and DA
→
→
→
respectively. Given that AS = ∼
a, AP = ∼
b and AJ = ∼c. State the
vectors for the following in terms of ∼
a, b or c.
→
→
→ ∼ ∼
→
(a) SD
(b) CJ
(c) RJ
(d) JQ
Solution
→
(a) SD = ∼
a
→
(b) CJ = – ∼c
→
(c) RJ = – ∼
a
D
~a
A
S
~c
R
Q
J
P
~b
C
B
→
(d) JQ = ∼
b
CHAPTER 8
Self Practice 8.2
1. By using suitable scales, draw and label the following vectors.
→
(a) XY represents 5 N of force to the right.
(b) RS represents 40 km of displacement to the southwest.
(c) ∼v represents velocity of 20 km h–1 to the west.
(d) a represents momentum of 7 kg m s–1 to the left.
2. The diagram on the right shows vector f that represents the force
∼
applied on an object from J to K. Find the magnitude and the
direction of vector f.
∼
K
N
f
~
J
1 cm
1 cm
3. Two cars, A and B are moving away from town O. Car A moves to the north while car B
moves to the east. Find the distance between the two cars after both cars travelled for one
→
→
hour, given that OA  = 90 km and OB  = 75 km.
216
8.1.2
Vectors
4. Find pairs of the same vectors in the diagram below.
M
a
~
N
~c
b
~
E
F
e
~
d G
~
f
~
C
H
5. The diagram on the right shows a regular hexagon ABCDEF.
(a) State the same vector for
→
→
→
(i) AB (ii) BC (iii) CD
(b) State the negative vector for
→
→
→
(i) AF (ii) FE (iii) ED
D
A
B
L
K
B
C
A
D
F
E
Making and verifying conjectures about the properties of scalar
multiplication on vectors
Observe the two following cases:
Case 1
~a
A
B
C
D
→
→
a, AB and CD in the
Observe vector ∼
diagram on the left.
→
It is found that AB = 2 × ∼
a or 2a
∼, and
→
a or 3a
CD = 3 × ∼
∼
Given that a
∼ = 5 units, thus
Case 2
~b
E
G
8.1.2
8.1.3
H
→
→
Observe vector ∼
b, EF and GH in the
diagram on the left.
→
It is found that EF = 2 × (–b
F
∼) or (–2)b
∼,
→ 1
1 b
and GH = × (–b
∼) or – 2 ∼
2
1 unit Given that b  = 8 units, thus
∼
→
→
1 unit
EF = 16 units and GH = 4 units.
( )
217
CHAPTER 8
→
1 unit →
AB = 10 units and CD  = 15 units.
1 unit
From case 1 and case 2, we can conclude that:
Multiplication of scalar k with vector ∼
a produces vector ka
∼, with the conditions:
(a) |ka
|
=
ka
.
∼
∼
(b) ka
is
in
the same direction with ∼
a if k > 0.
∼
(c) ka
is
in
the
opposite
direction
with
a if k < 0.
∼
∼
Example 5
State the following vectors in the diagram on the right in terms
of ∼c.
S
Q
c~
L
Solution
M
→
→
→
RS = 2c
∼, PQ = 3c
∼, LM = –2c
∼
R
Self Practice 8.3
Smart TIPS
1. State the following vectors in terms of ∼
a.
S
Q
~a
P
~x
y
~
R
P
Multiplication of vector
with scalar will also produce
vector quantity.
For example,
F = ma.
Force (vector)
= mass (scalar) × acceleration
(vector)
CHAPTER 8
Making and verifying conjectures about the parallel vectors
InQuirY 1
In groups
Aim: To create and verify conjectures about the relation between two parallel vectors
Instruction:
B
1. Consider the diagram on the right and answer the following
questions:
D
(a) Find the magnitude for each vector.
→ →
(b) Determine the ratio of |AB | : |CD |.
A
(c) Determine the gradient for straight line AB and CD. Are the
straight line AB and CD parallel?
→
→
(d) Express AB in term of CD .
2. Given two parallel vectors, a and b, what is the relation between
~
~
C
a and b? Discuss among your group members.
~
~
218
8.1.3
8.1.4
Vectors
From the results of Inquiry 1, we can conclude that if two vectors
are parallel, then one vector is a product of scalar with another
vector.
a and ∼
b are parallel if and only if ∼
a = kb
∼
∼, where k
is a constant
If ∼
a and ∼
b are two non-zero vectors and are not parallel, then
ha
=
kb
,
∼ ∼ thus h = k = 0.
MATHEMATICS
POCKET
Given three points, A, B and
C. The following are the
conditions for the points, to
be collinear.
→
→
(a) AB = kBC .
(b) AB parallel to BC.
(c) B is common point.
C
B
A
Example 6
→
→
→
→
Given PQ = ∼
a, QR = ∼
b, RS = –2a
∼ and ST = 4b
∼. Which pairs of vectors are parallel?
Solution
→
→
→
→
→
→
Given that PQ = ∼
a and RS = –2a
∼, then RS = –2PQ. Thus, PQ and RS are parallel.
→
→
→
→
→
→
Given that QR = ∼
b and ST = 4b
∼, then ST = 4QR. Thus, QR and ST are parallel.
Example 7
Solution
→
→
→
→
Given PQ = ∼
u and QR = 5u
∼, then, QR = 5PQ.
→
→
Thus, PQ and QR are parallel.
Since Q is a common point, P, Q and R
are collinear.
Mind Challenge
Given that points X, Y and
Z are collinear. Write the
→ →
relation between XY , XZ
→
and YZ .
CHAPTER 8
→
→
Given PQ = ∼
u and QR = 5u
∼, show that P, Q and R are collinear.
Example 8
Given that non-zero vectors, ∼
a and ∼
b are not parallel and (h – 1)a
∼ = (k + 5)b
∼, where h and k are
constant, find the value of h and k.
Solution
Given (h – 1) ∼
a = (k + 5)b
a and ∼
b are not parallel and non-zero, then
∼. When ∼
h – 1 = 0 and k + 5 = 0
h=1
k = −5
8.1.4
219
Self Practice 8.4
→
→
→
→ →
→
1. Given that AB = 5a
∼ and PQ = 20a
∼, express AB in terms of PQ if AB is parallel to PQ.
→
→
2. Show that point L, M and N are collinear given that LM = 6x
∼ and MN = 18x
∼.
3. Given that non-zero vector, ∼
u and ∼v are not parallel find the value of m and n for each of the
following.
(a) (4m + 3)u
(b) (m + n – 1)u
∼ = (n – 7)v
∼
∼ – (m – 2n – 10)v
∼=0
→
→
→
→
→
4. Given that XY and VW are parallel vectors, XY = 6 units and VW = 21 units, express VW in
→
terms of XY.
→
→
5. The points P, Q and R are collinear with PQ = ∼
a and QR = (k – 2)a
∼. Find the value of k if
→ 1→
PQ = PR , where k is a constant.
2
→
→
P
6. In the triangle PRS, QT and RS are two parallel vectors. Given
→
→
that PT : TS = 5 : 3, express SR in terms of QT .
Q
T
R
Intensive Practice 8.1
S
Scan the QR code or visit bit.ly/35mhddR for the quiz
1. The diagram on the right shows a trapezium ABCD. Given that
→
→
DC = ∼
u, AB = 6 cm and DC = 2 cm, write AB in terms of ∼
u.
A
B
CHAPTER 8
D
u
~
C
2. In the diagram on the right, AB and DC are parallel. Given that
C
D
→
→
→
E
DC = 1 AB and DC  = 4 cm.
3
→
(a) Find AB.
→
→
A
B
(b) If AE = 6a
and ED = 2b
, express
∼
∼
→
(i) EC in terms of ∼
a,
→
(ii) BE in terms of ∼
b.
→
→
3. Given that AB = 4x
∼ and AC = 6x
∼, show that A, B and C are collinear.
4. Vector ∼
a and vector ∼
b are non-zero and not parallel. Given that (h + k)a
∼ = (h – k + 1)b
∼ where
h and k are constant. Find the value of h and k.
→
→
5. Given that PQ = (k + 2) ∼x + 4y. If PQ is extended to point R with QR = hx
∼ + ∼y, express k in
∼
terms of h.
220
8.1.4
Vectors
8.2
Addition and Subtraction of Vectors
Performing addition and subtraction of vectors to produce resultant
vectors
InQuirY 2
In pairs
21st Century Learning
Aim: To identify the resultant vector
Instruction:
1. Observe the map on the right.
2. Dayang, Mia, Tan and Ranjit decided to
meet at the mini market.
3. Sketch the paths that can be taken by
them. Include the starting point and
the terminal point together with the
directions.
4. What can you say about the paths taken
by them?
Sketch of Ranjit's path
n
lta
u
es
r
cto
e
tv
Jalan SBBC7
From the results of Inquiry 2, it is found that the sketch of the
path taken by them produced a displacement, which is a resultant
vector. Resultant vector is the single vector that is produced from
addition of a few vectors.
R
The following are a few cases that involve resultant vector.
Jalan Sungai Limau
CHAPTER 8
Addition and subtraction of parallel vector
Case 1
A Addition of two parallel vector
+
4a
~
5a
~
=
9a
~
4a
∼ + 5a
∼ = 9a
∼
9a

=
4a

+
∼
∼ 5a
∼
B Subtraction of two parallel vector
7b
~
–
4b
~
3b
= ~
7b
∼ – 4b
∼ = 7b
∼ + (– 4b
∼) = 3b
∼
3b
∼ = 7b
∼ – 4b
∼
If vector ∼
a is parallel with vector ∼
b, then ∼
a–∼
b=∼
a + (–b
∼).
8.2.1
221
Case 2
BRAINSTORMING
Addition and subtraction of non-parallel vector
Does addition of vectors
follows commutative laws?
Discuss.
A Triangle Law
Triangle law for addition of two non-parallel vectors is given
→ → →
as AB + BC = AC .
b
~
B
a
~
C
a
~ + b~
A
This triangle law can be used for subtraction of two non-parallel vectors.
(–b)
~
a b
~– ~
a
~
a
~
a b
~–~
b
~
B Parallelogram Law
Two vectors, ∼
a and ∼
b initiated from the same point can be represented by two sides of
→
→
a parallelogram, AB and AD. Thus, the resultant vector of ∼
a and ∼
b is the diagonal of
→
parallelogram, AC.
B
a b
~ +~
a
~
CHAPTER 8
C
A
D
b
~
C Polygon Law
→ → → → →
Polygon law is given by AB + BC + CD + DE = AE.
a~
A
e~
222
E
B
d~
b~
D
C
c~
8.2.1
Vectors
Example 9
The diagram on the right shows a parallelogram PQRS.
S
(a) Express
→
→
→
(i) PQ in terms of PS and SQ ,
→
→
→
(ii) PR in terms of PQ and PS,
→
→
→
P
(iii) QR in terms of PR and PQ.
→
→
→
(b) Given that PQ = 2a
b and PS = 2b
a, express PR in terms of ∼
a and ∼
b.
∼+∼
∼–∼
R
Q
Solution
→ → →
→ → →
PQ = PS + SQ
(b) PR = PS + PQ
Triangle Law
→ → →
(ii) PR = PS + PQ = 2b
a + 2a
b
Parallelogram Law
∼–∼
∼+∼
→ → →
Triangle law
(iii) QR = QP + PR =
a + 3b
∼
∼
→ →
= –PQ + PR
→ →
= PR – PQ
(a) (i)
Example 10
The diagram on the right shows a pentagon ABCDE. Given that
→
→ →
→ →
→
→
BC = 1 AE, ED = 1 AB , AB = ∼
a and AE = ∼
b, express CD in
3
2
terms of ∼
a and ∼
b.
Solution
E
D
b~
C
A
a
~
B
CHAPTER 8
→
→ → → →
CD = CB + BA + AE + ED
=– 1 ∼
b–∼
a+∼
b+ 1 ∼
a
3
2
b– 1 ∼
a
= 2∼
3
2
Example 11
Hamzah rows his boat from point P across the river with
velocity, ∼v, 5 km h−1 to the north. The river stream flows with
velocity, ∼
a, 10 km h−1 to the east. The diagram on the right
shows the sketch movement of the boat and the river stream.
Calculate the new direction and velocity of the boat after
affected by the river stream.
8.2.1
10 km h–1
5 km h–1
P
223
Solution
Real velocity of the boat is ∼v + ∼
a.
2
2
5
+
10
v
+
a

=
!w
∼ ∼
= 11.18 km h−1
a = 10 km h–1
~
If θ is the angle formed by the north direction,
v = 5 km h–1
~
θ
then, tan θ = 10
5
θ = 63.43°
The boat is moving at the bearing 063.43° with the velocity of 11.18 km h−1.
v+a
~ ~
Self Practice 8.5
1. The diagram on the right shows vector ∼
u and vector ∼v.
Draw and label the resultant vector for each of the following:
(a) 2u
(b) 1 ∼v + 2u
∼ + ∼v
∼
2
3
(c) ∼
u – 2v
(d) 2u
∼
∼ – 2 ∼v
u~
v~
2. Vector p represents the velocity of 70 km h−1 to the south and vector q represent the velocity
∼
∼
of 80 km h−1 to the east. Find the direction and magnitude of the resultant vector, p + q.
∼ ∼
A
B
3. Given that ABCD is a trapezium with 3AB = 2DC.
Express the following in terms of ∼x and y.
∼
x~
→
→
(a) AB
(b) AC
→
→
D
C
y
(c) BC
(d) BD
CHAPTER 8
~
4. An airplane is flying to the north from airport P to airport Q for 1 200 km in 2 hours. The
wind blows from west with the velocity of 160 km h−1. Find
(a) the velocity of the plane without the influence of the wind,
(b) the original direction of the plane.
Solving problem involving vector
The problem involving addition and subtraction of vector for parallel vectors and non-parallel
vectors can be solved using triangle law, parallelogram law and polygon law.
Example 12
Mathematics
→
The position vector for three toy cars, A, B and C are OA = ∼
a+∼
b,
→
→
OB = 3a
∼ – 2b
∼ and OC = ha
∼ + 7b
∼, where h is a constant. Find the
value of h where these toy car A, B and C is placed in a straight line.
224
A
Application
B
C
O
8.2.1
8.2.2
8.1.1
Vectors
Solution
1 . Understanding the problem
◆
◆
→
→
→
Given OA = ∼
a+∼
b, OB = 3a
∼ – 2b
∼ and OC = ha
∼ + 7b
∼.
→
→
Toy cars A, B and C are located on a straight line, thus AC = kAB where k is a constant.
Find the value of k and h.
2 . Planning strategy
→
→
Find AB and AC using triangle law.
→
→
◆ Write the relation between AC = kAB .
◆ Find the value of k and h by comparing
the coefficient in the relation
→
→
AC = kAB .
◆
4 . Making a conclusion
When k = –2,
→
→
AC = kAB
= (–2)(2a
∼ – 3b
∼)
+
6b
= – 4a
∼
∼
When h = –3,
→
AC = (h – 1)a
∼ + 6b
∼
= (–3 – 1)a
∼ + 6b
∼
= – 4a
+
6b
∼
∼
3 . Implementing the strategy
→ → →
AB = AO + OB
→ →
= –OA + OB
= –∼
a–∼
b + 3a
∼ – 2b
∼
= 2a
∼ – 3b
∼
→ → →
AC = AO + OC
→ →
= –OA + OC
= –∼
a–∼
b + ha
∼ + 7b
∼
= (h – 1)a
∼ + 6b
∼
→
→
AC = kAB
(h – 1)a
∼ + 6b
∼ = k(2a
∼ – 3b
∼)
(h – 1)a
+
6b
=
(2k)a
–
(3k)b
∼
∼
∼
∼
Compare the coefficient of ∼
a and ∼
b,
h – 1 = 2k and 6 = –3k
h = 2k + 1
k = –2
Substitute k = –2 into h = 2k + 1,
h = 2(–2) + 1
= –3
CHAPTER 8
◆
Self Practice 8.6
→
→
1. Given O, X, Y and Z are four points with OX = 4x
– 2y, OY = kx
∼
∼ – ∼y and
∼
→
OZ = 6x
. If points X, Y and Z are collinear, find the value of k.
∼ + 5y
∼
2. The diagram on the right shows the plan of alleys of a residential area which
forms a quadrilateral ABCD. There is a lamp post at position E,
A
where BE : ED = 3 : 1. Alley AB and DC are parallel and DC = 4 AB.
3
→
→
(a) Express BD and AE in terms of ∼x and y.
24x~
∼
(b) Show that the alley AE is parallel to alley BC.
E
B
8.2.2
D
20y
~
C
225
Intensive Practice 8.2
Scan the QR code or visit bit.ly/2pWaGpX for the quiz
1. Express the following vectors in terms of ∼x and y.
∼
→
→
(b) QR
(a) AC
C
B
Q
y
~
D
A
x~
R
y
~
P
x~
P
y
~
P
y
~
y
~
CHAPTER 8
P
Q
x~
x~
Q
Q
→
→
3. In the diagram on the right, AB = ∼
a and AC = 4b
∼. Given Q is a
→
point on AC where AQ : QC = 1 : 3. Express BQ in terms of ∼
a
and ∼
b.
6. The diagram on the right shows a triangle OAB.
→
→
Given that OA = ∼
a, OB = ∼
b, 5BX = 3BA and OY : OA = 3 : 4.
(a) Find the following in terms of ∼
a and ∼
b.
→
→
(i) BA
(ii) BX
→
→
(iii) OX
(iv) BY
→
→
→
→
→
(b) Given that OP = λOX and BP = μBY. Express OP
in terms of
(i) λ, ∼
a and ∼
b,
(ii) μ, ∼
a and ∼
b,
(c) Hence, find the value of λ and μ.
a~
A
B
4b
~
4. Given that p = 2a
q = 4a
b and ∼r = ha
∼ + 3b
∼, ∼
∼–∼
∼ + (h + k)b
∼ where
∼
h and k are constants. Find the value of h and k if ∼r = 3p – 4q.
∼ ∼
5. The diagram on the right shows the sketch of a river. The width
of the river is 40 m and the velocity of the downstream river flow
is 1.8 m s–1. Hamid wanted to row his boat from P across the
river at R, but his boat was swept by the current flow and stopped
at Q in 12 seconds. Calculate the speed of Hamid's boat.
226
R
P
→
2. For the following diagrams, express vector PQ in terms of ∼x and y.
∼
(a)(b)(c)
Q
T
2x
~
x~
y
~
→
(c) PR
C
Q
R
40 m
1.8 m s–1
P 70 m
B
O
X
P
Y
A
Vectors
8.3
Vectors in Cartesian Plane
Representing vectors and determining the magnitude of vectors in
the Cartesian plane
21st Century Learning
( )
12
KIMANIS
PENAMPANG
6
PAPAR
4
2
KUDAT
KOTA
MARUDU
RANAU
BELURAN
AU
8
TUARAN
G
10
KOTA
BELUD
IN
Arding wishes to explore a province in Sabah. Arding
is at a place located at coordinate (1, 3). Next, he
moves 5 units parallel to x-axis and 4 units parallel to
y-axis to a location in another province. He promises to
meet his friend, Timan at the location. Timan moves at
translation 2 from his place to meet Arding.
6
y
EN
Aim: To identify the resultant vector
Instruction:
1. Observe the map of Sabah that is drawn on a
Cartesian plane on the right.
2. Observe the following situation carefully:
K
In groups
KINABATANGAN
TENOM
InQuirY 3
SILAM
PENSIANGAN
KALABAKAN
From the results of Inquiry 3, a vector can be expressed as the combination of parallel vector
and non-parallel vector. On the Cartesian plane, vector will be expressed as the combination of
parallel vector with x-axis and y-axis.
Vector with magnitude 1 unit and parallel with x-axis is called
vector ∼i and is written as ∼i = 1 , ∼i  = 1.
0
()
y
j
~
O
8.3.1
~i
x
Vector with magnitude 1 unit and parallel with y-axis is called
vector j and is written as j = 0 ,  j  = 1.
∼
∼
1 ∼
()
227
CHAPTER 8
x
3. Plot on the Cartesian plane, the movement and the
0
4
6
8
2
position of Arding and Timan.
4. What is the name of the province where they both meet?
5. Express the translation of the movement of Arding from the location of the first
province to the location of the second province.
6. Find the distance, in units, between Arding's first location and Timan's first location
with their meeting point.
7. Present the findings in front of the class and host a question and answer session with
your friends.
Observe the following diagram:
•
•
B
•
The coordinates of point B is B(x, y).
→
The position vector of point B relative to point O is OB.
→
OB can be written as the combination of vector ∼i and j,
∼
which is xi∼ + yj.
∼
→
• OB can be written in the form of column vector, x .
y
→
• Magnitude of OB = !w
x2 + y2
y
x
O
()
Example 13
Given that points A(1, 2), B(– 4, 5), C(8, –3), D(–7, – 4) and O is the origin on a Cartesian
→ → →
→
plane. Express the vectors OA, OB, OC and OD in the forms of
(a) x
(b) xi∼ + yj
∼
y
()
Solution
()
( )
( )
( )
→
→
→
→
–7
(a) OA = 1 , OB = – 4 , OC = 8 , OD =
–4
2
5
–3
→
→
→
→
(b) OA = ∼i + 2j, OB = – 4i∼ + 5j, OC = 8i∼ – 3j, OD = –7i∼ – 4 j
∼
∼
∼
∼
CHAPTER 8
Example 14
The diagram on the right shows vector ∼
a, ∼
b, ∼c, ∼
d and ∼e on a
Cartesian plane.
(a) Express each vector in the forms of xi∼ + yj and x .
∼
y
(b) Find the magnitude for every vector.
(c) Are vectors ∼
b and ∼e parallel? Give your reason.
()
Solution
()
( )
( )
( )
( )
(a) ∼
a = 3i∼ + 2j, ∼
a = 3 , b
,b= 4
∼ = 4i∼ – 2j
∼
∼ ∼ –2
2
,c= 0
, d
d = –3
∼c = –3j
∼ = –3i∼, ∼
∼ ∼ –3
0
, e = –4
∼e = – 4i∼ + 3j
∼ ∼
3
228
y
a~
e~
b~
c~
x
d~
8.3.1
Vectors
(b) a
32 + 22 , b
42 + (–2)2
∼ = !w
∼ = !w
= 3.606 units
= 4.472 units
∼c = !w
02 + (–3)2 , d
(–3)2 + 02
∼ = !w
= 3 units
= 3 units
∼e = !w
(– 4)2 + 32
= 5 units
(c) Vectors ∼
b and ∼e are not parallel because ∼
b � ke
b � gradient ∼e.
∼ or the gradient ∼
Self Practice 8.7
1. The diagram below shows 6 points on a Cartesian plane.
y
C
–4
–2
F
–8 –6
–4
–2
D
O
B
A
2
E
4
6
8
10
12
14
x
→ → → → →
→
Express OA, OF, BC , FA, DE and DO in the form of
(a) xi∼ + yj,
∼
(b) column vector.
2. Given that point A(−2 , 3), point B(5, 8) and O which is the origin on a Cartesian plan.
(a) Find the position vector of point B.
→
(b) Calculate AB.
( ) ( )
B
A
E
D
C
()
4. Given that p = 3 , q = –5 and ∼r = 1 are representing the position vector of point P, Q and
∼ – 4 ∼ –7
5
R.
(a) Write the vectors p, q and ∼r in the form of xi∼ + yj.
∼ ∼
∼
(b) State the coordinates of points P, Q and R.
(c) Calculate the length of vector p, q and ∼r.
∼ ∼
8.3.1
229
CHAPTER 8
3. The diagram on the right shows 5 points, A, B, C, D and E on a grid.
(a) Express the following vectors in the form of resultant vector
of vectors ∼i and j.
∼
→
→
(ii) BA
(i) AB
→
→
(iii) BC
(iv) DC
→
→
(v) AC
(vi) DE
(b) State the pair of vectors that are parallel and explain your reason.
(c) State the pair of vectors that are negative and give your reason.
Describing and determining the unit vector in a direction of a vector
You have learned that ∼i and j are the unit vectors which are parallel to positive x-axis and
∼
y-axis respectively. Let's find out about the unit vector in the direction of vector that is not
parallel to the x-axis or y-axis.
InQuirY 4
In pairs
21st Century Learning
Aim: To determine the unit vectors in the direction of a vector given
Instruction:
1. Scan the QR code or visit the link on the right.
bit.ly/2Is5lNN
2. Drag the slider x1 and y1 to view the changes of the unit
vector on Cartesian plane and the calculated unit vector.
3. Compare the unit vector with each of the changes in x1 and y1 values.
4. Discuss the formula used to find the unit vector in direction with a vector.
From the results of Inquiry 4, unit vector in the direction of a vector
can be found by dividing vector with the magnitude of the vector.
In general:
If r = xi∼ + yj, then vector unit in direction ∼r is
∼
xi + yj
r
r̂∼ = ∼ = ∼ ∼ .
∼r !w
x2 + y2
MATHEMATICS
POCKET
Unit vector is a vector in
the direction of a vector
with magnitude of 1 unit.
Example 15
CHAPTER 8
Given point A(4, 3), find the unit vector in the direction of vector
→
OA. Express the answer in the forms of
(b) column vector.
(a) component ∼i and j,
∼
Solution
→
(a) OA = ∼
a = 4i∼ + 3j
∼
2
2
a

=
4
+
3
!w
∼
= 5 units
4i + 3j
Unit vector in the component ∼i and j is ∼
âa = ∼ ∼.
∼
5
(b) Unit vector in the form of column vector is
âa = 1 4
∼
5 3
4
5
= 3 or 0.8
0.6
5
()
() ( )
230
Quick COUNT
Determine the magnitude of
vector 4∼i + 3 j using scientific
∼
calculator.
1. Press MENU
2. Press 1
3. Press SHIFT +
4. Screen will show:
Pol (
5. Press 4 SHIFT )
3 =
6. Screen will show:
Pol (4, 3)
r=5
8.3.2
Vectors
Example 16
Given that – 1 ∼i + kj is a unit vector, find the value of k.
∼
3
Solution
!(– 13 ) + k = 1
! 19 + k = 1
2
2
Magnitude of the unit vector is 1
2
1 + k2 = 1
9
k2 = 8
9
k = ±0.9428
Self Practice 8.8
()
1. Calculate the magnitude for the following vectors.
0
(a) 3 (b) – 4 (c) – 4 (d) –12i∼ – 5j (e) 6i
∼
2
–7
7
()
( )
2. Find the unit vector in the direction of the following vectors.
(a) 3i∼ + 2j
(b) –i∼ – 9j
(c) 4
∼
∼
0
()
()
(d)
–8
(–15
)
3. Determine whether the following vectors are unit vectors.
– 1
2
!w
7
!w
(c) – 0.6 (d) 7 ∼i + 24 j (e) 2 ∼i + j
(a) 0 (b)
3
1
∼
–1
– 0.8
25
25∼
3
2
!w
( )
4. Find the value of k for the following unit vectors.
(b) k
(a) 0
k
0
()
()
(d) k
k
()
(e) 0.5i∼ + kj
∼
(c)
( 1k )
CHAPTER 8
( )
(f) ki∼ + 13 j
84∼
pi + 8j
5. Given the unit vector in the direction of vector ∼
u is û∼ = ∼ ∼, find the possible values of p.
73
!w
6. Given û∼ = (1 – k)i∼ + hj, express h in terms of k.
∼
8.3.2
231
Performing arithmetic operations on two or more vectors
A Addition of two or more vectors
()
()
() ()
( )
Consider ∼
a = a1 and ∼
b = b1 .
a2
b2
a+∼
b = a1 + b1
∼
a2
b2
a1 + b1
=
a2 + b2
Then, ∼
a+∼
b = (a1∼i + a2 j) + (b1 ∼i + b2 j)
∼
∼
= (a1 + b1)∼i + (a2 + b2)j
∼
b
~
a + ~b
~
~a
b2
a2 + b2
a2 b1
a1
a1 + b1
Gather component ~i and j , then sum
~
up separately
Example 17
Find the addition of the following vectors.
a= 4 ,∼
(a) ∼
b = 0 and ∼c = –1
1
–3
6
() ( )
Solution
( )
() ( ) ( )
()
(a) ∼
a+∼
b + ∼c = 4 + 0 + –1
1
–3
6
= 3
4
(b) ∼v = 3i∼ + 2j and w
∼ = 4i∼ – 5j
∼
∼
(b) ∼v + w
) + (4i∼ – 5j)
∼ = (3i∼ + 2j
∼
∼
= (3 + 4)i∼ + (2 – 5)j
∼
= 7i∼ – 3j
∼
B Subtraction of two vectors
CHAPTER 8
The same method for the addition of vectors can be used in the operation of subtraction of two
vectors.
Example 18
Find p – q for the following pairs of vectors.
∼ ∼
(a) p = 7 and q = 4
∼ –1
∼
1
( )
Solution
( ) ()
( )
( )
(a) p – q = 7 – 4
∼ ∼ –1
1
7
–
4
=
–1 – 1
= 3
–2
232
()
(b) p = 2i∼ – j and q = 3i∼ + 5j
∼
∼
∼
∼
(b) p – q = (2i∼ – j) – (3i∼ + 5j)
∼ ∼
∼
∼
= (2 – 3)i∼ + (–1 – 5)j
∼
= –i∼ – 6j
∼
8.3.3
Vectors
C Multiplication of vector with scalar
When a vector is multiplied with a scalar, both component ∼i and j are also multiplied with the
∼
scalar.
Example 19
For each of the following vectors, find
–4
(a) –3s
(b) 2r
.
∼, given ∼s = 2 ,
∼, given ∼r = 5i∼ – 3j
∼
( )
Solution
( )
( )
–4
(a) –3s
∼ = –3 2
= 12
–6
(b) 2r
)
∼ = 2(5i∼ – 3j
∼
= 10i∼ – 6 j
∼
Smart TIPS
Arithmetic operation
involving parallel vector is
performed using the same
method as non-parallel
vector.
D Combination of arithmetic operation on vectors
Combination of arithmetic operation applied on vectors need to follow the operation rules of
mathematics. Multiplication with scalar need to be performed before addition and subtraction.
Example 20
( ) ( )
()
Given p = 6 , q = – 4 and ∼r = 7 , determine the vector 3p + q – 2r
.
∼ –3 ∼
∼ ∼ ∼
5
8
Solution
( ) ( ) ()
( ) ( ) ( )
( )
Self Practice 8.9
( ) ( )
CHAPTER 8
3p + q – 2r
= 3 6 + –4 – 2 7
∼ ∼ ∼
–3
5
8
18
–
4
14
=
+
–
–9
5
16
0
=
–20
()
1. Given ∼
a = –3 , ∼
b = 4 and ∼c = 1 , find
5
–12
8
(a) 2a
b + ∼c
∼–∼
(b) –3a
∼ + 2b
∼ – ∼c
(c) 1 ∼
b + ∼c – 3a
∼
2
2. Given ∼
u = 3i∼ + 6j, ∼v = –2i∼ – 8j and w
, find
∼ = 3i∼ – 4j
∼
∼
∼
(a) ∼
(c) 1 ∼v + w
u – 2v
(b) 3u
∼ – 3u
∼
∼+w
∼
∼ + 2v
∼–w
∼
2
8.3.3
(d) 1 ∼
b–∼
a + 3c
∼
4
(d) 1 ∼v – w
∼ + 3u
∼
4
233
Solving problems involving vectors
By applying the knowledge learnt, problems involving vectors can be solved easily, especially
the problem involving our daily lives.
Example 21
Mathematics
Application
One particle is moving from point A(5, 10) with the velocity vector (3i∼ – j) m s−1. After t seconds
∼
→ →
leaving A, the particle is on point S, with OS = OA + tv
.
Find
the
speed
and the position of the
∼
particle from O after 4 seconds. When will the particle reside on the right side of origin O?
Solution
3 . Implementing the strategy
1 . Understanding the problem
◆
Original position vector,
→
OA = ∼
a = 5i∼ + 10j = 5 .
∼ 10
( )
y
( )
Velocity vector, ∼v = 3i∼ – j = 3 .
∼ –1
◆ Speed is the magnitude of velocity
vector.
◆ The particle is on the right side of O if
component j in the position vector
∼
is zero.
◆
2 . Planning the strategy
Find v
∼ to determine the speed.
Find position of particle after
→ →
4 seconds using OS = OA + tv
∼ or
a + ∼vt when t = 4.
∼s = ∼
◆ The particle is on the right side of O
when the y component in ∼s = x is
y
zero.
CHAPTER 8
◆
◆
()
4 . Making a conclusion
Distance AM = !w
302 + 102
= !w
1 000 m
1 000
!w
10
= !w
10 m s–1
Then, speed =
234
10 m
a~
A(5, 10)
t v~
S
s~
O 5m B
3t m
M
x
Speed, v
32 + (–1)2
∼ = !w
= !w
10 m s–1
After 4 seconds, ∼s = ∼
a + 4v
∼,
s= 5 +4 3
10
–1
17
=
6
The particle is at point (17, 6).
Position vector after t seconds,
5
3
∼s = 10 + t –1
= 5 + 3t
10 – t
The position of the particle after t
seconds is
→
OS = ∼s = 5 + 3t .
10 – t
The particle is at the right side of
origin O when
y=0
10 – t = 0
t = 10 seconds
( ) ( )
( )
( ) ( )
( )
(
)
8.3.4
Vectors
Self Practice 8.10
1. A toy car is at point A(−3, −2). The car is then moved with a constant velocity (2i∼ – 3j) cm s−1.
∼
Find the position vector of the toy car after 2.5 seconds.
( )
2. The position vector of boat A, t hours after leaving the port O is t 30 while the position
15
vector of boat B is 50 + t 10 . Determine the velocity of boat A and boat B. Will the two
5
10
boats meet?
( ) ( )
Intensive Practice 8.3
( )
Scan the QR code or visit bit.ly/2MqRicw for the quiz
()
1. Two forces F1 = – 4 and F2 = 7 are applied on an object as the
3
5
diagram on the right.
(a) Find the resultant force.
(b) Calculate the magnitude of the resultant force.
F2
F1
2. Given p = (k – 3)i∼ + 14j and q = ∼i + (k – 8)j with k is a constant. If p is parallel to q, find the
∼
∼
∼
∼
∼
∼
value of k.
() ( )
( )
b = 5 and ∼c = m . If ∼
u is parallel to ∼v, find
3. Given ∼
u=∼
b–∼
a and ∼v = ∼c – ∼
b, with ∼
a= 3 ,∼
1
–2
–6
the value of m. After that find u
∼ : v
∼.
( )
5. A swimmer swims at a velocity ∼v = 2.4 . It is found that the stream flows at a velocity of
1.5
0.5
. Find the magnitude and direction of the resultant velocity of the swimmer.
a=
∼
–2.1
( )
6. Given ∼r = 2i∼ – 5j and ∼s = mi∼ – 3j, find the value of m if
∼
∼
(a) ∼r + ∼s = 10,
(b) ∼r parallel to ∼s,
(c) (2r
∼ – ∼s) is parallel to the y-axis.
7. Given
()
k
1 is a unit vector, find the value of k.
2
!w
( )
8. The length of vector ∼v is 5 units and the direction is opposite with vector 2 , find vector ∼v.
–1
8.3.4
235
CHAPTER 8
→
→
4. Given the triangle ABC with AB = 2i∼ – j and AC = 10i∼ + 5j. R is a point on BC with
∼
∼
→
→
BR = 1 BC . Find
2
→
(a) BC ,
→
(b) the unit vector in the direction of BC ,
→
(c) AR .
9. Vector p = (m – 1)i∼ + 2j is orthogonal with vector q = 8i∼ + nj. Express m in terms of n.
∼
∼
∼
∼
10. Ship M left port O when the sea was calm, with the velocity of vM = 6i∼ + 8j km h–1. At the
∼
same time, ship N left port Q with velocity of vN = 4i∼ + 4j km h–1. Given that position vector
∼
→
of port Q is OQ = 50i∼ + 20j.
∼
→
(a) After t hours, position vector of ship M is OM = t(6i∼ + 8j). Find the position vector of
∼
ship N at that time.
(b) Show that ship M will cross ship N and find the time when this occurs.
Summary OF Chapter 8
•
•
�o�
a�
�
�
•
→
AB = a~
→
Negative vector of AB
→ →
is –AB or BA
Zero vector, 0~
r��
r�
�e
Unit vector in direction of
r = xi + yj is
~ ~ ~
xi + yj
^r = ~ ~
~ x2 + y2
Magnitude,
| r | = x2 + y2
~
��n��
o��p
•
•
xi + yj
~ ~
x with
y
()
()
()
CHAPTER 8
i= 1
~ 0 and
0
j=
1
~
��e��t�
�n
�t��
v�
ka
~ is a vector that is parallel
VECTORS
to ~
a and the magnitude is
a |.
k × |~
Multiplication of vector:
k × a~ = ka
~
• Addition of vectors
→
→
→
2AB + 3AB = 5AB
• Subtraction of vectors
4a
~ – a~ = 3a~
WRITE YOUR JOURNAL
In pairs, find the differences between a scalar quantity and a vector quantity. Compare the
methods used to perform arithmetic operations for both quantities. After that, find information
from the internet regarding the application of vectors in your daily life. Write a report and
discuss your findings.
236
Vectors
MASTERY PRACTICE
1. The diagram on the right shows three vectors, ∼
a, ∼
b and ∼c
that are not parallel. Express PL1
→
(a) XY in terms of ∼
a and ∼
b,
→
(b) XZ in terms of ∼
b and ∼c.
a
∼
Y
b
∼
Z
∼c
X
→
→
→
→
2. Given PQ = 3ka
∼ – 4b
∼ and XY = 4a
∼ + 8b
∼. If PQ parallel to XY, find the value of k.
3. Given p = mi∼ – nj is a unit vector in the direction of p, express m in terms of n.
∼
∼
∼
4. Given ∼
u = ki∼ + hj and ∼v = ∼i – 4j. If u
k2 + h2, express h in terms of k.
∼ + ∼v = !w
∼
∼
( )
PL2
PL2
PL2
6. The diagram on the right shows triangle PQR with
→
→
→ →
PQ = 3i∼ and PR = 2j. Given RS : SQ = 2 : 3,
∼
→
express RS in terms of ∼i and j. PL3
∼
R
S
2j
∼
P
7. The diagram on the right shows a trapezium BCDE
→
→
→
→
with DC = ∼
u and EB = ∼v. If ED = 1 BC ,
2
→
express BC in terms of ∼
u and ∼v. PL3
C
u~
D
v~
E
8. The diagram on the right shows a regular hexagon,
→
→
ABCDEF with centre O. Given FA = ∼
a and FB = ∼
b,
(a) express the following in terms of ∼
a and/or ∼
b,
→
→
(i) AB
(ii) FO
→
→
(iv) BC
(v) FD
→
→
(b) state the relationship between AB and FC ,
→
→
(c) determine whether AC and FD are parallel.
Q
3∼i
PL3
→
(iii) FC
→
(vi) AD
F
E
a~
B
A
b~
O
B
C
D
237
CHAPTER 8
( )
→
→
5. Given A(3, 4), AB = 5 and BC = 10 . Find
12
–3
→
(a) the unit vector in the direction of AC,
(b) the coordinates C.
PL2
9. The position vector of city A is –10i∼ + 10j and position vector of city B is 10i∼ – 11j. The
∼
∼
position of city A, B and C are collinear and the distance between city A and city C is two
times the distance between city A and city B. The distance between cities is measured in
kilometer. Find PL4
→
(a) vector AB,
(b) the distance between city A and city B,
→
(c) vector OC .
A
10. The diagram on the right shows a quadrilateral OABC.
M is the midpoint of AC and OM : OB = 2 : 3. Given
→
→
→
OA = 3u
∼ + 2v
∼, OC = 9u
∼ + 2v
∼ and CB = 3kv
∼, where k
is constant, PL4
(a) express in terms of ∼
u and/or ∼v,
→
→
(i) AC
(ii) OM
→
(b) express OB in terms of
(i) ∼
u and ∼v,
(ii) ∼
u, ∼v and k.
Hence, find the value of k.
3u
~ + 2v
~
(
238
)
CHAPTER 8
)
3kv
~
C
Y
11. The diagram on the right shows roads of a housing area that form
a rectangle OABC. Building D is at OB road and building E is at
OA road. Given OD = 3 OB and OE : OA = 1 : 2.
4
C
Building Y is at AB road which is extended with BY = 1 AB.
2
OA road is represented by vector 4a
∼ while OC road is
PL5
represented by vector 4c
.
∼
(a) Express vector that represents the following road in
terms of ∼
a and ∼c.
O
→
→
→
→
(i) OB (ii) OD (iii) OY (iv) ED
(b) Prove that building E, D and Y reside on the same straight line.
(
M
9u
~ + 2v
~
O
12. The diagram on the right shows the positions and directions of
the boats which belong to Arul, Ben and Raju in a solar boat
competition. Arul’s and Ben’s boats move in the same direction of
1
–1
the stream. The velocity of the stream is given by w
∼ = ∼i + 3∼j m s ,
while the velocity of Arul’s boat is ∼
a = (3i∼ + j) m s–1 and the
∼
velocity of Ben’s boat is ∼
b = (6i∼ + 2j) m s–1. PL5
∼
(a) Calculate the resultant velocity of Arul’s boat and the
resultant velocity of Ben’s boat. After that, find the
difference between the speed of the two boats.
(b) Raju’s boat is deviated from the path. Given that the
4
velocity of Raju’s boat is ∼r = 2i∼ – j m s–1. Find the unit
3∼
vector in the direction of the resultant velocity of the boat.
B
D
E
Arul
B
A
Ben
Raju
Starting line
Vectors
Exploring
MATHEMATICS
Madam Tan is a housewife who visits several locations every day. The diagram below shows
the displacement vector ∼
b, ∼c, ∼e and m
∼ that represent the journey of Madam Tan from her house
in A to the location she normally visits.
~c
A
D
m
~
Guide:
A : Madam Tan's
house
B : Market
C : Mother’s house
D : School
M : Kindergarten
X : Grocery store
~e
b
~
M
X
m
~
B
C
()
Arithmetic Resultant
operation
vector
(a)
m
∼ – ∼e
(b)
b
∼
m
∼– 2
(c)
(d)
(e)
b – ∼c
∼
b
∼
2
∼c + ∼e + m
∼
→
AD
Arithmetic Resultant
operation
vector
b
∼
(f)
∼c – 2
b
∼c – ∼
(g)
2
(h)
(i)
(j)
∼c – 2m
∼
b
b + ∼c – ∼
∼
2
b – 2m
∼
∼
Arithmetic
operation
(k)
m
b
∼ – ∼c – ∼
(l)
b – ∼c
∼
2
Resultant
vector
(m) ∼
b + ∼c – m
∼ – ∼e
b + ∼c
∼
(n)
2
(o)
b – ∼c
∼c + ∼
239
CHAPTER 8
x
Write the vectors ∼
b, ∼c, ∼e and m
. [1 side = 1 km].
∼ in the forms of y and xi∼ + yj
∼
1. After that, find the shortest distance from Madam Tan's house to every location according
to the given displacement vector.
2. Madam Tan will send her son to the kindergarten before sending her daughter to the
→
→
→
school. Observe that the resultant vector AD = AM + MD = m
∼ – ∼e obeys the triangle law.
State the other resultant vectors that obey the:
(a) Triangle law,
(b) Parallelogram law,
(c) Polygon law
3. Copy and complete the table by filling up the resultant vectors that are represented by
combinations of vectors through arithmetic operations of the vectors.
Solution of
Triangles
CHAPTER
9
What will be learnt?
Sine Rule
Cosine Rule
Area of a Triangle
Application of Sine Rule, Cosine Rule and
Area of a Triangle
List of
Learning
Standards
bit.ly/2pfbIgk
CHAPTER
BAB 99
wORDS
KEY W
Acute angle
Obtuse angle
Sine rule
Cosine rule
Ambiguous case
Included angle
Non-included angle
Three dimension
240
Sudut tirus
Sudut cakah
Petua sinus
Petua kosinus
Kes berambiguiti
Sudut kandung
Sudut bukan kandung
Tiga matra
9.1.1
Solution of Triangles
Did you
The architecture in the shape of triangles
appears to be very unique in a building. This
shape is also used to decorate the walls of
building to portray attractive and modern
images. We are captivated by the uniqueness
of the triangular shapes of this architecture.
However, how can we determine the height
of the architecture? What is the information
needed to measure the area of each triangle?
Know?
Abu Wafa Muhammad Ibn Muhammad Ibn
Yahya Ibn Ismail Buzhani (940-997 M) was
a Persian astronomer and mathematician.
Abu Wafa learnt trigonometry in Iraq in the
year of 959 and developed a few important
theories especially in the field of geometry
and trigonometry.
For further information:
bit.ly/2q4x80j
SIGNIFICANCE
OF THIS CH
APTER
CHAPTER 9
There are various fields which use triangles to
solve problems. For example:
Astronomy field uses the concept of triangle
to measure the distance between stars.
Geography field uses the solution of triangles
to measure distance beween various places.
Satellite field uses triangles in its navigation
system.
Scan the QR code
to watch the video in
Masbro village, Melaka.
bit.ly/2VjgB3F
9.1.1
241
9.1
Sine Rule
Making and verifying conjectures on the relationship between the
ratio of length of sides of a triangle with the sine of the opposite
angles
In our daily lives, we often face situations involving triangles.
For example, the solution to find the height of a kite. When it
involves a non-right-angled triangle, Pythagoras Theorem is
unsuitable to be used. There are other methods to find the solution
of non-right-angled triangles. Let’s explore.
37°
INQUIRY 1
90 m
62°
In pairs
Aim: To make conjectures on the relationship between the ratios of length of sides of a
triangle with the sine of the opposite angles.
Instructions:
1. Copy or print the table below.
2. Complete the following table based on the given triangles.
a
sin A
Triangle
b
sin B
c
sin C
(a) Acute-angled triangle
A
7.46 cm 78.93°
B
9.8 cm
59.89°
41.18°
11.12 cm
C
(b) Obtuse-angled triangle
A
CHAPTER
BAB 99
46.3°
4.06 cm
10.785 cm
113.5°
C
8.5 cm
20.2°
B
3. Discuss in pairs and state the conjecture on the relationship between the ratio of
length of sides of a triangle with the sine of the opposite angles.
242
9.1.1
Solution of Triangles
From the results of Inquiry 1, it is found that
a = b = c
sin A sin B sin C
or
sin A = sin B = sin C
a
b
c
Is this conjecture valid for all types of acute-angled triangles and obtuse-angled triangles? Let’s
explore.
Diagram (a) and Diagram (b) is an acute-angled triangle and an obtuse-angled triangle
respectively. CD is perpendicular to AB and it is represented by h.
C
a
B
h
C
b
h
A
D
Diagram (a) Acute-angled triangle
D
b
a
B
c
A
Diagram (b) Obtuse-angled triangle
Consider triangle BCD,
h = sin B
a
Then, h = a sin B … 1
QR
The relationship between
the ratio of length of sides
of a triangle with the sine
of the opposite angles by
using GeoGebra software.
Consider triangle ACD,
h = sin A
b
Then, h = b sin A … 2
1 = 2 , a sin B = b sin A
a = b
sin A sin B
sin A = sin B
or
a
b
bit.ly/2p1IUb2
Sine rule
For any triangle ABC,
a = b = c
or
sin A sin B sin C
9.1.1
sin A = sin B = sin C
a
b
c
Mind Challenge
What do you get if the sine
rule is used in right-angled
triangles?
243
CHAPTER 9
It is observed that for any acute-angled triangle and
obtuse-angled triangle, the ratio of length of sides with the sine of
the opposite angles are the same. This relationship is known as the
sine rule.
QR
Example 1
Write the sine rule that relates the sides and angles of the
following triangles.
(a)
(b)
H
F
k
d
e
j
E
f
D
Revision of solution of
right-angled triangles.
J
h
bit.ly/2B0Crjo
K
Mind Challenge
Solution
(a)
d = e = f
sin D sin E sin F
(b)
h
j
=
= k
sin H sin J sin K
Discuss with your friends and
prove that
a = c
sin A sin C
Self Practice 9.1
1. Write the sine rule that relates the sides and angles of the following triangles.
(a)(b)(c)
q
K
P
R
l
r
p
M
Q
P
8 cm
m
k
L
6 cm
120° 40°
R
Q
Solving triangles involving sine rule
Solving a triangle means finding the measurements such as length of sides, size of angles,
perimeter or area of the triangle. We can solve problems involving triangles by using sine rule.
CHAPTER
BAB 99
Example 2
Find the value of x in the following triangles.
(a)
R
(b)
Z
x
x
14.3 cm
48°
P
244
80° Q
6.2 cm
X 55.5°
7.05 cm
Y
9.1.1
9.1.2
Solution of Triangles
Solution
x = 14.3
(b) sin x = sin 55.5°
sin 48° sin 80°
7.05
6.2
14.3
sin
55.5° × 7.05
× sin 48°sin x =
x =
sin 80°
6.2
= 10.791 cm= 0.9371
x = 69.57°
(a)
Example 3
FLASHBACK
The diagram below shows the positions of Qistina’s house,
Ron’s house and a post office.
N
N
Post office
Qistina’s house
Q
12 km
N
30° A
9.4 km
R
O
Ron’s house
Calculate
(a) the bearing of the post office from Qistina’s house,
(b) the bearing of the post office from Ron’s house,
(c) the distance from Qistina’s house to the post office.
Solution
Assume the positions of the post office, Qistina’s house and
Ron’s house are represented by P, Q and R respectively.
sin Q
(a) sin 98° =
9.4
12
sin
98° × 9.4
sin Q =
12
= 0.7757
∠Q = 50.87°
Bearing P from Q = 90° – 50.87°
= 39.13°
Thus, the bearing of the post office from Qistina’s house is 039.13°.
(b) ∠R = 180° – ∠P – ∠Q
= 180° – 98° – 50.87°
= 31.13°
Bearing P from R = 270° + 31.13°
= 301.13°
Thus, the bearing of the post office from Ron’s house is 301.13°.
9.1.2
Bearing A from O in the
above diagram is written as
030° or N30°E.
Smart TIPS
In order to solve a triangle
by using the sine rule, the
following conditions must
be learnt:
(a) Two angles and length
of one side, or
(b) The lengths of two sides
and a non-included
angle.
Mind Challenge
What is a non-included
angle? Explain.
245
CHAPTER 9
98°
In Geography, bearing
is used to indicate the
direction of a certain place
from a reference point. For
example
r represents the distance from
r
= 12
Qistina’s house to the post office
sin 31.13° sin 98°
12
× sin 31.13°
r=
sin 98°
= 6.265
(c)
Thus, the distance from Qistina’s house to the post office is 6.265 km.
Self Practice 9.2
1. Determine the value of m in the following triangles.
(a)(b)(c)
N
M
55°
L
78°
6.7 cm
N 40° m
L
M
m
L
8 cm
6.5 cm
2. The diagram on the right shows a tower which is
inclined by 7° from the vertical line. At a distance
of 100 m from the side of the tower, the angle of
elevation is 30.5°. Estimate the height, XY,
in m, of the tower.
12.4 cm
m
7°
43° M
115°
N
X
30.5°
Y
Determining the existence and solving triangles involving
ambiguous cases
The diagram below shows two triangles, ABC and AB1C with the
lengths of two sides and a non-included angle given as follow:
C
CHAPTER
BAB 99
6 cm
A
30°
4 cm
B1
4 cm
MATHEMATICS
POCKET
Ambiguity means
inexactness or the quality
of being open to more than
one interpretation.
B
Based on the diagram above, it is observed that two different triangles can be constructed
by using the given non-included angle and lengths of two sides. The two triangles can be
constructed by using the same set of information given. This is known as ambiguous case.
246
9.1.2
9.1.3
9.1.4
Solution of Triangles
INQUIRY 2
In groups
21st Century Learning
Aim: To determine the conditions for the existence of ambiguous case
Instructions:
1. Scan the QR code or visit the link given.
2. Given ∠BAC = 45°, length of side, c = 10 cm and h is the height
bit.ly/33c6SiC
of the triangle.
3. Drag the slider a to the left and to the right. Observe the changes that take place.
4. Discuss in groups and answer the following questions:
(a) State your observations when
(i) a < h
(ii) a = h
(iii) a > h
(vi) a > c
(iv) a < c
(v) a = c
(b) Does ambiguous case exist?
5. Each group appoints a representative to do the presentation in the class.
6. Students from other groups are encouraged to ask questions.
7. The teacher will summarise all the presentations.
From the results of Inquiry 2, there are three conditions for the existence of triangles as shown
in the following:
B
No triangle exists
a
h
c
A
a<h
C
B
c
A
A
C
a=h
a
c
a=h
C
ac
B
B
c
Two triangles exist
A
h
h<a<c
c
a
C
A
a
h
C
h<a<c
Ambiguous case exists if:
(a) Given the lengths of two sides, a and c, and a non-included angle, ∠A which is acute.
(b) The side which is opposite the non-included angle, a is shorter than the other side, c, but it
is longer than the height, h, of the triangle.
9.1.3
9.1.4
247
CHAPTER 9
One triangle exists
B
Example 4
Determine whether ambiguous case exists for the following triangles. Explain your answer.
(a)
(b)
R
R
7 cm
P
30°
21 cm
4 cm
Q
80°
Q
17 cm
P
40°
Solution
(a) Yes, ambiguous case exists in triangle PQR because the
non-included angle ∠QPR = 30° and the side RQ is shorter
than the side PR but it is longer than the height of triangle.
(b) Ambiguous case does not exist because the angles of the
two sides are given.
R
7 cm
P
30°
4 cm
Q
Q1
Example 5
In the triangle ABC, ∠BAC = 40°, AB = 20 cm and BC = 14 cm. Calculate the possible values
of ∠C and ∠B.
Solution
Determine whether ambiguous case exists for triangle ABC.
Height, h = 20 sin 40°
= 12.856 cm
Since h < BC < AB, then ambiguous case exists.
CHAPTER
BAB 99
Look at the sketch of the triangle ABC in the diagram on the right. A
The two triangles that exist are ABC and ABC1.
For triangle ABC,
sin ∠C = sin 40°
20
14
20
sin 40°
sin ∠C =
14
= 0.9183
∠C = 66.68°
∠B = 180° – 40° – 66.68°
= 73.32°
248
14 cm B
20 cm
h
40°
14 cm
C
C1
∠C1 = 180° – 66.68°
= 113.32°
∠B1 = 180° – 40° – 113.32°
= 26.68°
9.1.3
9.1.4
Solution of Triangles
Self Practice 9.3
1. For each of the following triangles, determine whether ambiguous case exists.
(a) ΔABC; ∠B = 62.5°, BC = 14.5 cm and AC = 10 cm.
(b) ΔPQR; ∠R = 28°, QR = 8.2 cm and PQ = 11.4 cm.
2. The diagram on the right shows an incomplete triangle PQR.
PQ = 15.5 cm and ∠QPR = 35°.
Given QR = 10.5 cm,
(a) find the possible values of ∠QRP,
(b) hence, find the possible lengths of PR.
Q
15.5 cm
P
35°
R
Solving problems related to triangles using the sine rule
Example 6
Azyan and Christine stand straight in front of a flag
post as shown in the diagram. The elevation angle of
top of the pole from Azyan is 36° whereas the elevation
angle of top of the pole from Christine is 50°. Badrul
is standing on the left side of the flag pole and the
elevation angle of the top of the pole from him is the
same as Christine. The distance between Azyan and
Christine is 35 m. Find the distance between Azyan and
Badrul if the height of three of them are the same.
Solution
Azyan Badrul
Christine
Represent the positions of Azyan, Badrul, Christine and the top of the pole with A, B, C and D
respectively.
9.1.3
9.1.4
9.1.5
50°
C
CHAPTER 9
∠ADC = 180° – 50° – 36°
D
= 94°
∠BDC = 180° – 50° – 50°
= 80°
DC = 35
50°
A 36°
sin 36° sin 94°
B
35 m
DC = 35 × sin 36°
sin 94°
= 20.6227 m
BC = 20.6227
sin 80° sin 50°
BC = 20.6227 × sin 80°
sin 50°
= 26.5120 m
AB = 35 m – 26.5120 m
= 8.488 m
Then, the distance between Azyan and Badrul is 8.488 m.
249
Self Practice 9.4
1. Encik Samad makes a plan for his vegetable
farm as shown in the diagram. Encik Samad
wants to install two water sprinklers in the
middle of the farm. The water tap which
controls the water sprinklers is installed at one
corner of the farm. The distance between the
two water sprinklers is 6 m and the distance
between the water tap and the nearest water
sprinkler is 5 m. The angle formed between the
tap and both water sprinklers is 25°. Calculate
the distance between the tap and the furthest
water sprinkler.
2. A group of scouts organised an activity on
crossing a river during a motivational camp.
They tied a rope from tree P to tree Q and tree
R on the other side of the river as shown in the
diagram. The distance between tree Q and tree
R is 80 m and a 50° angle is formed between
tree Q and tree R at P. Find the value of x, the
distance from tree P to tree Q.
Intensive Practice 9.1
Water tap
Crop area
Crop area
Water
sprinkler
Water
sprinkler
Q
80 m
P
xm
100 m
Scan the QR code or visit bit.ly/35oe3pQ for the quiz
1. The diagram on the right shows a triangle ABC such that
∠B = 77°, ∠C = 39° and AC = 40.5 cm. Calculate the value of
∠A, a and c.
A
77°
39°
a
C
B
10 cm
cm
17 cm
10
2. The diagram on the right shows a triangle ABD. Point C and
point E lie on the straight line AD.
(a) Find the lengths of BE, CE and DE.
(b) Calculate ∠EAB, ∠BCE, ∠BCD, ∠ABD and
∠CBD.
(c) Explain the ambiguous case in the diagram
D
on the right.
40.5 cm
c
B
CHAPTER
BAB 99
R
C
E
6 cm
A
3 cm and ∠QPR = 40°.
3. In the obtuse-angled triangle PQR, PR = 14 cm, QR = 6!w
(a) State the obtuse angle and find the size of that angle.
(b) Calculate the length of PQ.
250
9.1.5
9.1.1
Solution of Triangles
4. The diagram on the right shows a square picture frame which
is hung by Amira using two pieces of ropes. Amira finds that
the picture frame inclines towards the right. The angle formed
between the longer rope and the frame is 48°. The lengths of the
rope are 20 cm and 15 cm respectively. Calculate the perimeter
of the frame.
5. The diagram on the right shows the position of Puan
Azizah’s house and the houses of her two children, Amir
and Anita. Another child, Aida wants to build her house
such that all the three houses of the siblings are collinear
and the distance from her house and Anita’s house to
Puan Azizah’s house is the same. Find the distance
between Anita’s house and Aida’s house.
9.2
48°
Puan Azizah’s
house
250 m
120°
Amir’s
house
Anita’s
150 m house
Cosine Rule
Observe the diagrams below.
Q
B
A
132°
35 m
C
25 cm
12 cm
42 m
P
30 cm
R
Cosine rule
For any triangle ABC,
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
9.2.1
CHAPTER 9
How do you determine the length of AB and angle PQR? Can both triangles be solved by using
the sine rule?
When the lengths of two sides and an included angle or the lengths of three sides are given,
the triangle cannot be solved by using the sine rule. The triangles with such conditions can be
solved by using the cosine rule.
C
b
A
a
c
B
251
Verifying the cosine rule
Is the cosine rule true for all types of triangles? Let’s explore.
Consider the triangle ABC in the diagram. By using the Pythagoras Theorem in the
triangle ACD,
b2 = h2 + (a – x)2
A
b2 = h2 + a2 – 2ax + x2 … 1
c
Use the Pythagoras Theorem in the triangle ABD,
c2 = h2 + x2
… 2
h2 = c2 – x2
Substitute 2 into 1 .
b2 = c2 – x2 + a2 – 2ax + x2
…
b2 = c2 + a2 – 2ax
B
x
a–x
D
C
3
In the triangle ABD,
cos B = x
c
x = c cos B
Substitute x = c cos B into
b2 = c2 + a2 – 2ac cos B
b
h
Mind Challenge
Can the cosine rule be
used on the right-angled
triangles? Explain.
3
.
This equation is one of the formulae of cosine rule. Try to verify the cosine rule for
obtuse-angled triangle.
Solving triangles involving the cosine rule
Cosine rule can be used to find the length or unknown angle in a triangle when the lengths of
two sides and an included angle or the lengths of three sides are given.
Example 7
In the triangle ABC, AC = 21 cm, BC = 15 cm and ∠C = 35°. Find the length of AB.
CHAPTER
BAB 99
Solution
Sketch the triangle ABC.
By using the cosine rule,
x2 = 152 + 212 – 2(15)(21) kos 35°
= 225 + 441 – 630 kos 35°
= 149.9342
149.9342
Then, x = !w
= 12.245 cm
252
A
21 cm
x
B
35°
15 cm
C
9.2.1
9.2.2
Solution of Triangles
Example 8
The diagram on the right shows a triangle
JKL with the length of JK = 30 cm,
KL = 25 cm and JL = 35 cm. Find the J
value of ∠KJL.
Solution
30 cm
K
Smart TIPS
25 cm
To find the angles, the
formulae of cosine rule can
be written as follow:
2
2
2
• cos A = b + c – a
2bc
2
2
2
• cos B = a + c – b
2ac
2
2
2
• cos C = a + b – c
2ab
35 cm
L
By using the cosine rule,
252 = 302 + 352 – 2(30)(35) cos ∠KJL
2
2
2
cos ∠KJL = 30 + 35 – 25
2(30)(35)
= 0.7143
∠KJL = 44.41°
Example 9
In the diagram on the right, QST and PSR are straight lines.
Find the length of QR.
Solution
R
T
10 cm
9 cm
By using the cosine rule,
92 = 62 + 62 – 2(6)(6) cos ∠PST
2
2
2
cos ∠PST = 6 + 6 – 9
2(6)(6)
= – 0.1250
∠PST = 97.18°
S
6
P
cm
4 cm
Q
QR2 = 42 + 102 – 2(4)(10) cos 97.18°
= 125.999
QR = 11.225 cm
Self Practice 9.5
1. Find the value of x in the following triangles.
(a)
(b)
45°
5 cm
x
3 cm
F
9.2.2
G
(c)
J
55°13ʹ
Q
x
9 cm
K
12 cm
CHAPTER 9
E
x
75 m
L
P
32°
100 m
R
253
2. Find the value of θ in the following triangles.
(a)
(b)
E
J
20 cm
F
15.7 cm
G
14 cm
(c)
10.8 cm
L
7 cm
P
10 cm
9 cm
12 cm
K
Q
3. The diagram on the right shows a quadrilateral PQRS.
Find the angle PQR.
Q
5 cm
R
6 cm
R
9 cm
8.7 cm
P
12.5 cm
42.3°
S
Solving problems involving the cosine rule
Example 10
Mr. Sivaraja has a plot of land in the shape of
trapezium PQRS as shown in the diagram on the
right. He puts up a fence around the land.There is
a tree at a distance of 50 m from the vertex Q.
Mr. Sivaraja wants to divide the land into two parts
by putting some additional fence from the vertex S
to the tree. Calculate the length of fence put up by
Mr. Sivaraja.
65 m
S
R
75 m
85 m
P
Q
105 m
CHAPTER
BAB 99
Solution
SQ = !w
652 + 752
= 99.2472 m
99.24722 = 852 + 1052 – 2(85)(105) cos ∠SPQ
2
2
2
cos ∠SPQ = 85 + 105 – 99.2472
2(85)(105)
∠SPQ = 61.93°
ST 2 = 552 + 852 – 2(55)(85) cos 61.93°
= 5850.3581
ST = 76.488 m
The length of the additional fence is 76.488 m.
254
65 m
S
R
85 m
P
55 m
75 m
T
Q
50 m
9.2.2
9.2.3
Solution of Triangles
Self Practice 9.6
1. Farid carried out archery training on a field. The
diagram on the right shows two target boards which
have to be struck by arrows. The distance between
Farid and board I and board II are 25 m and 45 m
respectively. The standing position of the shooting
is 38° between board I and board II. Calculate the
distance between board I and board II.
Board I
Board II
25 m
45 m
38°
2. Frank planted four iron rods in the ground and installed wires to
build clotheslines. The sketch of the clotheslines built by Frank
is shown in the diagram on the right. The wire AB is parallel
to the wire DC. Calculate the total length of the wire used by
Frank.
D 5m
8m
10 m
A
3. The diagram on the right shows the positions of the house of
four friends, Amin, Imran, Johan and Raden. During Hari Raya,
Amin wants to visit all the three houses of his friends. Amin
intends to pick up Imran and then send him back home before
he returns to his house. What is the total distance travelled by
Amin for the whole journey?
C
B
10 m
Johan
8 km
Imran
9 km
120°
Amin
9.5 km
Raden
Scan the QR code or visit bit.ly/2VvtloA for the quiz
1. A card is in the shape of a parallelogram. Given the lengths of the diagonals of the card are
6 cm and 10 cm respectively. The acute angle between the diagonals are 62°. Calculate the
lengths of sides of the card.
N
K
km
L
15
2. The diagram on the right shows the positions of three towns, J, K
and L. Given the bearing of K from J is 020° and bearing of L
from J is 055°, find the distance between town K and town L.
20 km
J
9.2.3
255
CHAPTER 9
Intensive Practice 9.2
3. Bunga Raya ship left a port and sailed east for a distance of 28 km. Bunga Orkid ship left the
same port and sailed for 49 km. If the final distance between the two ships is 36 km, find the
angle between the routes of Bunga Raya ship and Bunga Orkid ship.
4. The diagram on the right shows a pond in the shape of triangle MNP.
Given cos θ = 4 , MP = 8 m, PQ = 7 m and QN = 4 m. Encik Raja
5
decorates the pond by arranging stones around the pond.
P
Calculate the length of stones arranged by Encik Raja around
the pond.
9.3
M
8m
7m Q 4m
N
Area of a Triangle
Deriving the formula and determining the area of a triangle
The diagram on the right shows the shape of a window
of a building and it is in the shape of a triangle. What
is the information required to calculate the area of the
window in the diagram and what is the formula that
you will use to determine the area of the window?
You have learnt that the area of the triangle can be
determined by using the following formula:
C
Height
A
Base
B
CHAPTER
BAB 99
Area = 1 × base × height
2
FLASHBACK
C
The formula for the area of triangle can be used when
the length of base and height of triangle are given. How
do you find the area of a triangle without knowing the
length of the base and the height? Let’s explore the
method used to derive the formula for the area
of a triangle.
256
a
B
h
c
b
A
h = a sin B
h = b sin A
9.3.1
Solution of Triangles
INQUIRY 3
In groups
21st Century Learning
Aim: To derive the formula for the area of a triangle
Instructions:
1. Start this activity in pairs.
2. Observe the following shapes of triangles.
P
B
c
A
h
a
r
C
b
Triangle I
h
X
h
x
R
p
Q
z
Y
q
Triangle II
y
Z
Triangle III
3. Find the height of each of the triangles by using trigonometric ratios.
4. Then, copy and complete the following table based on the triangles above.
Triangle
Base
I
AC
Height
Area
II
III
5. Compare the formulae for the area of the three triangles and state the conclusion
based on your findings.
6. Form a few groups. Then, each pair will share the results and conclusions in their
respective group.
From the results of Inquiry 3, if the lengths of two sides and an included angle are only given,
the area of a triangle can be calculated by using the following formulae:
c
A
a
b
C
Area = 1 ab sin C
2
= 1 ac sin B
2
= 1 bc sin A
2
QR
Methods to derive the
formulae for the area of a
triangle.
bit.ly/316kXNk
9.3.1
257
CHAPTER 9
B
Example 11
Find the area of triangle JKL in the diagram on the right.
K
Solution
Included angle = 69°
Area = 1 (6)(4.4) sin 69°
2
= 12.323 cm2
69°
6 cm
4.4 cm
J
L
Example 12
The area of triangle DEF is 50 cm2. Given DE = 8.6 cm, DF = 14.2 cm and ∠EDF = θ, find
the value of θ.
Solution
E
1 (8.6)(14.2) sin θ = 50
2
sin θ = 50
61.06
θ = 54.97°
8.6 cm
D
F
14.2 cm
Self Practice 9.7
1. Find the area of the following triangles.
(a)(b)(c)
C
Q
16.2 cm
A
49°
7 cm
18.4 cm
B
Z
125°
P
10 cm
10 cm
R
X
CHAPTER
BAB 99
2. In the diagram on the right, the area of triangle LMN is
78.72 cm2. Find the length of LM.
60°
35°
N
17 cm
20°
L
3. The diagram on the right shows triangle BCD and triangle ABD.
Find the area of triangle ABD.
A
Y
M
B
20 cm
24.18°
10 cm
D
55°
C
4. Find the area of triangle XYZ, given x = 5.5 m, z = 7 m and ∠Y = 70°30ʹ.
258
9.3.1
Solution of Triangles
Determining the area of a triangle using the Heron’s formula
Consider the following triangle ABC:
B
10 cm
A
5 cm
C
11 cm
When only the lengths of each side are given, the area of the
triangle can be determined by using the Heron’s formula.
The solving steps are as follows:
Step 1
Calculate the semi perimeter, s = a + b + c,
2
such that a, b and c are the lengths of sides.
Step 2
Substitute the values of s, a, b and c into the
following formula:
s(s – a)(s – b)(s – c)
Area = !w
Mathematics Museum
Hero of Alexandria also
known as Heron is a Greek
mathematician. Heron’s
formula was named after
him and had been written in
his book entitled “Metrica”.
Example 13
Find the area of the triangle below.
C
12.3 cm
A
5.5 cm
9.6 cm
B
QR
Verification of Heron’s
formula.
s = 5.5 + 9.6 + 12.3
2
= 13.7
Area = !13.7(13.7 – 5.5)(13.7 – 9.6)(13.7 – 12.3)
= 25.39 cm2
9.3.2
CHAPTER 9
Solution
bit.ly/2WrkvbM
259
Example 14
A group of scouts planted three pieces of wood
in a camping ground to build a fire pit. A rope of
length 22 m is used to tie around those woods
as shown in the diagram. The rope formed an
isosceles triangle. The length of the rope at the
side of equal length is 8 m. Calculate the area of
the region for them to build the fire pit.
8m
8m
Solution
C
8m
8m
A
c
B
Given the perimeter of triangle = 22 m, a = 8 m, b = 8 m.
c = 22 – 8 – 8
=6m
Alternative Method
62 = 82 + 82 – 2(8)(8)cos∠ACB
∠ACB = 44.05°
Area = 1 (8)(8) sin 44.05°
2
= 22.249 m2
s = 22
2
= 11
Area = !11(11 – 8)(11 – 8)(11 – 6)
= 22.249 m2
Thus, the area of the region for building the fire pit is 22.249 m2.
Self Practice 9.8
CHAPTER
BAB 99
1. The diagram on the right shows a triangle ABC such
that AB = 5.4 cm, AC = 6.1 cm and BC = 7.3 cm.
Calculate the area, in cm2, of the triangle ABC.
A
5.4 cm
B
2. The diagram on the right shows two triangles, EFJ
and EGH. EFG and EJH are straight line. Calculate
the area, in cm2, of the shaded region.
5 cm
C
7.3 cm
6 cm
E
260
6.1 cm
H
J
9 cm
F 3 cm
G
9.3.2
Solution of Triangles
3. Mr. Sammy wants to paint the wall of his room. He draws
a triangular shape on the wall and he will paint the triangle
with green paint. The shape of the triangle is as shown in the
diagram on the right. The lengths of sides of the triangle are 2x
135 cm2. Find
cm, 3x cm and 4x cm respectively. The area is !w
the value of x.
P
3x cm
4x cm
R
2x cm
Q
Solving problems involving areas of triangles
Example 15
A
The diagram on the right shows the plan of a plot of agricultural
60 m
land in the shape of triangle ABC owned by Encik Munzir. The
50 m
Q
part APQ will be planted with chillies and the remaining part
will be planted with cabbage. Given AP = 50 m, AQ = 60 m,
P
AB = 80 m, AC = 130 m and BC = 140 m, find the area of land 30 m
B
which will be planted with cabbage.
140 m
C
Solution
Assume L1 as the area of triangle ABC and L2 as the area of triangle APQ.
Alternative Method
2
2
2
cos A = 80 + 130 – 140
2(80)(130)
A = 79.75°
CHAPTER 9
Heron’s formula to find L1.
s = 80 + 130 + 140
2
= 175
L1 = !175(175 – 80)(175 – 130)(175 – 140)
= 5117.0670 m2
Use the formula, area = 1 bc sin A to find ∠BAC.
2
1 (80)(130) sin ∠BAC = 5117.0670
2
sin ∠BAC = 5117.0670
1 (80)(130)
2
∠BAC = 79.75°
Use the formula = 1 pq sin A to find L2.
2
1
L2 = (50)(60) sin 79.75°
2
= 1476.0610 m2
Thus, the area of land which will be planted with cabbage = L1 – L2
= 5117.0670 – 1476.0610
= 3641.006 m2
9.3.2
9.3.3
261
Self Practice 9.9
1. Mr. Khan won a tender to install carpet in an office. Calculate
the area of carpet required to fill up the office space as shown
in the diagram on the right.
16 m
18 m
11.5 m
20.5 m
2. The diagram on the right shows a decoration in the shape of
pyramid. The decoration has a triangular base PQR.
Vertex V is vertically above vertex P. Given PQ = 4 cm,
PV = 10 cm, VR = 15 cm and ∠VQR = 80°, calculate the
area of the inclined surface of the decoration.
V
P
Q
Intensive Practice 9.3
R
Scan QR code or visit bit.ly/2ohScA1 for the quiz
1. The diagram on the right shows a triangle ABC.
A
Given the area of triangle ABC = 18 cm2 and sin θ = 2 , find
3
(a) the length of AC,
(b) the area of triangle ABD.
B
3 cm
D
θ
6 cm
C
2. A regular pentagon has sides of 5 cm each. Find the area of the regular pentagon.
3. Mei Ling wants to prepare a greeting card in the shape of a triangle. The area of the card is
30 cm2 and the lengths of two sides are 8 cm and 11 cm. Find the possible length of the third
side.
4. The length of the sides of a triangle are 3x cm, (x – 1) cm and (3x + 1) cm. Given the
perimeter of the triangle is 63 cm. Calculate the area, in cm2, of the triangle.
CHAPTER
BAB 99
5. Pooja fenced up a plot of land in the shape as shown in the
diagram on the right. Given BD = 5 m, BC = 7 m, CD = 8 m
and AE = 12 m. BDE and ADC are straight lines. Calculate
the area of land fenced up by Pooja.
A
B
C
6. The diagram on the right shows a decoration rack in the shape
of a triangle, EFG. Given FG = 15 cm, EG = 16 cm
and EF = 17 cm, find the height of the rack.
F
262
40° E
D
E
G
9.3.3
Solution of Triangles
9.4
Application of Sine Rule, Cosine Rule and Area of a Triangle
Solving problems involving triangles
Example 16
MATHEMATICS APPLICATION
Mr. Tan wants to paint the roof of the garage. The diagram on the right
is a sketch of the front view of the roof of the garage. He found out that
the wood on one part of the roof is longer than the wood on the other
part of the roof.
(a) Calculate the length of wood on the longer part of the roof and the
distance between both the walls of the garage.
(b) What is the area of the front roof in the shape of triangle, in m2,
which will be painted by Mr. Tan?
5.2 m
30°
50°
Solution
◆
◆
◆
Length of one side of the roof = 5.2 m.
Two angles given 30° and 50°.
Calculate the length of two sides and
the area of triangle.
2 . Planning a strategy
Draw the triangle ABC which
represents the front view of the roof
of the garage.
◆ Length of one side of the roof, AC = y is
calculated using sine rule.
◆ Determine ∠BAC and hence calculate
BC using cosine rule.
◆ Find the area of triangle ABC using the
formula:
Area = 1 ab sin C
2
or Heron’s formula.
◆
3 . Implementing the strategy
(a)
A
ym
C
30°
5.2 m
50°
B
By using sine rule,
y
= 5.2
sin 50° sin 30°
y = 5.2 × sin 50°
sin 30°
= 7.967 m
Thus, the length of the other part of
the roof is 7.967 m.
∠BAC = 180° – 30° – 50°
= 100°
By using cosine rule,
BC2 = 5.22 + 7.9672 – 2(5.2)(7.967)
cos 100°
BC = 10.24 m
Thus, the distance between both walls
of the garage is 10.24 m.
(b) Area of triangle ABC
= 1 (5.2)(10.24) sin 50°
2
= 20.40 m2
Thus, the triangular area which will be
painted by Mr. Tan is 20.40 m2.
263
9.4.1
CHAPTER 9
1 . Understanding the problem
4 . Making a conclusion
Using Heron’s formula,
s = 5.2 + 7.967 + 10.24 = 11.7035 m
2
Area
= !11.7035(11.7035 – 5.2)(11.7035 – 7.967)(11.7035 – 10.24)
≈ 20.40 m2
The value of AC, BC and the area calculated are valid.
Example 17
The diagram on the right shows a glass prism and the sketch
of the prism. The cross section of the prism is an equilateral
triangle sides of 6 cm and the height of the prism is 8 cm.
Calculate
(a) the angle between BD and CD,
(b) the area of BCD,
(c) the angle between the plane BCD and the vertical plane
BCEF.
E
D
F
C
A
B
CHAPTER
BAB 99
Solution
(a) CD = !w
62 + 82
= 10 cm
62 = 102 + 102 – 2(10)(10) cos ∠BDC
2
2
2
cos ∠BDC = 10 + 10 – 6
2(10)(10)
∠BDC = 34.92°
Thus, the angle between BD and CD is 34.92°.
(b) Area of triangle BCD = 1 (10)(10) sin 34.92°
2
= 28.622 cm2
(c) Based on the diagram on the right, the angle between the
plane BCD and vertical plane BCEF is ∠DGH.
62 – 32
DH = !w
= 5.1962
tan ∠DGH = 5.1962
8
∠DGH = 33°
D
10 cm
B
10 cm
6 cm
C
E
D
F
H
C
A
G
B
264
9.4.1
Solution of Triangles
Self Practice 9.10
1. In an examination hall, the tables of Daniel, Darvin and Cindy
are at positions A, B and C respectively. These three points form
a triangle as shown in the diagram on the right. The distance
between Daniel’s table and Cindy’s table is 10 m, Daniel’s
table and Darvin’s table is 4.027 m whereas Darvin’s table
A
and Cindy’s table is 6.575 m. Prove that the sum of the
interior angles of the triangle formed is 180°.
C
10 m
B
4.027 m
2. The diagram on the right shows a children’s toy in the shape of
a cone and its upper portion is cut off. The surfaces in the shape
of circles, centre O and centre P are horizontal and the OP axis
is vertical. There is a straight line that joins A to C. Given
OA = 5 cm, PB = 10 cm, OP = 16 cm and ∠BPC = 90°,
calculate
(a) the length of AC,
(b) the area of plane ABC.
3. The positions of two towns, A and B, are shown in the
Cartesian plane in the diagram on the right. Find the angle
between the position vector of town A and town B relative to
the origin O. Hence, find the area of the region in the shape
of the triangle OAB.
6.575 m
A 5 cm O
16 cm
P
B 10 cm
90°
C
y
A(–4, 3)
B(1, 1)
x
O
Scan QR code or visit bit.ly/2IG2l0m for the quiz
N
4 cm Q
W
W
P
R
10 cm
N
13 km
10 km
040°
265
CHAPTER 9
5 cm
S
9.4.1
6 cm
T
E
2. The diagram on the right shows the position of an oil rig, a
tanker and a helicopter. The bearing of the helicopter from
the tanker is 40°. Given the distance between the helicopter
and the tanker is 13 km whereas the distance between the
helicopter and the oil rig is 10 km. Calculate the distance, in
km, between the tanker and the oil rig.
S
S
1. The diagram on the right shows the front view of a doll
house built by Melly. The coloured part is the veranda roof
of the doll house. PTS and PQR are straight lines.
(a) Calculate the area of the roof QRST.
(b) Point U lies on PR such that SU = SR,
calculate ∠SUP.
E
Intensive Practice 9.4
3. The diagram on the right shows a gift box in the shape of a cuboid. Q
(a) Calculate the area of the plane ACQ.
P
(b) Hence, state another plane which has the same area as
the plane ACQ.
B
A
R
6 cm
S
8 cm
C
4 cm
D
4. A ship sailed for 20 km to Port Bentara at the bearing of 120° from Port Astaka. Then, the
ship sailed for 30 km to Port Cindai at the bearing of 225° from Port Bentara. Calculate the
distance and the bearing of Port Cindai from Port Astaka.
5. The elevation angle of the peak of a mountain from Arman is 20°. Then, Arman walks
horizontally towards the mountain that is 800 m away and the elevation angle becomes 45°.
Estimate the height of the mountain from the level of Arman’s position.
Summary OF Chapter 9
CHAPTER
BAB 99
Ambiguous case
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
��s
��i
�d
Given two sides and
an included angle
Given lengths of
three sides
1
L = –– ab sin C
2
1
L = –– ac sin B
2
1
L = –– bc sin A
2
Heron’s formula
�
• Given lengths of two sides
and an included angle
• Given lengths of three sides
c�
��a �
c��d
��i � � s
Cosine rule
Solve problems
involving triangle
266
v
��
��
f�� m � � a
a
b
c
––––– = ––––– = –––––
sin A sin B
sin C
�v
�� �
f�� m
f��m��a�
• Given two angles and
length of one side
• Given lengths of two
sides and a non-included
angle
Area of triangle
g
Sine rule
f�� m � � a�
c��d��i
��
s
SOLUTION OF TRIANGLES
L = s(s – a)(s – b)(s – c)
a+b+c
where s = ––––––––
2
Solution of Triangles
WRITE YOUR JOURNAL
1. Draw a flow chart which shows the steps used in choosing the suitable rules to find
(a) length of sides or size of angles of a triangle,
(b) the area of a triangle.
2. Surf the internet to get
(a) the examples of usage of sine rule, cosine rule and the formula of the area of triangles
in our daily lives,
(b) the area of the Kuala Lumpur Golden Triangle, India Golden Triangle and Bermuda
Triangle.
MASTERY PRACTICE
1. (a) Given ∠ABC = 50°, ∠BAC = 72° and c = 5.8 cm, calculate the length of a and b.
(b) Given the sides of triangle PQR are p = 8.28 cm, q = 6.56 cm and r = 3.63 cm,
find ∠P, ∠Q and ∠R. PL2
3 cm
X
9 cm
55°13ʹ
S
14 cm
Y
31°52ʹ
x cm
6 cm
x cm
Z
Q
77°
R
3. The diagram on the right shows a right-angled triangle ABC.
Point D lies on AB. Calculate PL3
(a) the length of AC,
(b) the area of triangle ADC.
A
7 cm
D
3 cm
B
4. Given triangle XYZ such that ∠X = 42.2°, x = 10 cm and z = 13.4 cm.
(a) Sketch two possible shapes of the triangle.
(b) Hence, find the possible values of ∠Z.
(c) Calculate the area of triangle XYZ for the obtuse angle of ∠Z.
5. The diagram on the right shows five points, A, B, C, D
and E which forms quadrilaterals. BCD is a straight line,
∠ACB is obtuse and the area of triangle ADE
is 20 cm2. Calculate PL4
B
(a) the length of AD,
(b) ∠DAE.
m
C
PL4
9 cm
30°
10 c
A
10 cm
E
5 cm
C
6 cm
D
267
CHAPTER 9
2. Find the value of x in each of the following diagrams. PL3
(a)
(b)
P
PL1
6. In the diagram on the right, PQR is an equilateral triangle with sides
of 6 cm and lies on a horizontal plane. Point T is 4 cm vertically
above the midpoint of PQ. Calculate PL5
(a) the angle formed by TR and triangle PQR,
(b) the area of plane TPR.
T
Q
P
7. A group of girl scouts took part in a camping. They set up three
tents with the positions as shown in the diagram on the right.
The positions of the three tents formed a triangle ABC. PL5
(a) Calculate the obtuse angle ACB.
(b) Draw and label another triangle other than triangle ABC
which shows the possible position of tent C such that the
distance AB and AC and ∠ABC remain unchanged.
(c) Tent C has to be relocated to other position but the distance
between tent A and tent B and angle BAC formed between
the tents remain unchanged. Calculate the distance AC
such that only one triangle can be formed.
8. The diagram on the right shows a glass block in the shape of a
pyramid VABC. The base of the block is an isosceles triangle and
AB = AC = 5.2 cm. V is the vertex of the block such that
BV = CV = 3 cm. The angle between the inclined plane VBC and
the base ABC is 50°. Calculate PL5
(a) ∠BAC, given the base area is 8.69 cm2,
(b) the length of AV, given the angle between the line AV and
the base is 25°,
(c) the surface area of VAB of the glass block.
6 cm
6 cm
R
B
8m
A
7m
50°
C
V
C
A
B
CHAPTER
BAB 99
9. Rashid drove a boat westwards. He noticed a lighthouse at a distance of 25 km away at the
bearing of 235°. PL5
(a) Sketch a diagram to illustrate the above situation.
(b) What is the distance travelled by the boat if its distance from the lighthouse is 16 km?
(c) Rashid continued to drive the boat until his distance from the lighthouse is 16 km again.
(i) Calculate the distance between the first position and the second position of the boat.
(ii) What is the bearing of the lighthouse from the boat when the boat is at the second
position?
10. The diagram on the right shows the positions of four petrol stations,
J, K, L and M in a district. Given the distance JK = 40 km,
KL = 80 km, LM = 65 km and ∠JKL = 44°. PL5
M
(a) Calculate
65 km
(i) the distance JL,
(ii) ∠JML,
J
(iii) the area of KLM.
40 km
(b) Without doing calculations, determine the petrol
station which is the furthest from petrol station K.
L
K
80 km
Explain.
(c) If a car travels along the road KL, calculate the shortest distance of the car from petrol
station M.
268
Solution of Triangles
E
11. Mary coloured the three triangles, ABC, ACD and CED
B
5 cm
such that ACE and BCD are straight lines. Given
C
∠DCE = 50.05° and ∠CED is obtuse. PL6
(a) Calculate
7 cm
(i) ∠CED,
9 cm
(ii) the length of AB,
(iii) the area of triangle AED.
(b) The straight line AB is extended to point Bʹ such A
that CBʹ = CB. On the same diagram, draw and colour the triangle BCBʹ.
12. In the diagram on the right, WYZ is a straight line.
Given sin ∠XYW = 10. PL6
11
(a) Find sin ∠YXZ.
(b) Calculate the area of triangle XYZ. Hence, find the length of XW.
(c) State two situations so that the ambiguous case exists in the
triangle on the right.
Exploring
6.5 cm
D
X
12 cm
W
Y
Z
4 cm
MATHEMATICS
You are given a roll of wire of 100 metres length. You have to fence up a
region in the shape of an isosceles triangle. The diagram on the right
shows the sketch of the triangular region.
A
c
B
b
a
C
(a) Complete the following table to find the possible lengths of sides, a, b and c, of the
triangle that can be formed by using the wire.
b
49
48
c
49
48
Area of triangle
CHAPTER 9
a
2
4
(b) By using suitable formulae and technology, calculate the area of each triangle.
(c) Hence, predict the maximum area which can be fenced up and state the shape of the
triangle.
269
CHAPTER
Index Numbers
10
What will be learnt?
Index Numbers
Composite Index
List of
Learning
Standards
bit.ly/2q4n8El
w ORDS
KEY W
Nombor indeks
Indeks harga
Kuantiti pada
masa asas
Kuantiti pada
masa tertentu
Indeks gubahan
Pemberat
CHAPTER 10
Index number
Price index
Quantity at
base time
Quantity at
specific time
Composite index
Weightage
270
10.1.1
Index Numbers
Did you Know?
In the year 1764, Giovanni Rinaldo Carli
(1720-1795) who was an Italian economist
calculated the price ratios of three goods from
the year 1500 to 1750. The average of the
price ratios of the three goods represented the
measures of change that had happened in the
period of 250 years. His idea has resulted in
the extended usage of index number till today.
For further information:
bit.ly/33ngFCU
SIGNIFICANCE
OF THIS CH
APTER
Consumer Price Index (CPI) measures
the changes in the prices of goods and
services that represent the average
purchasing pattern of a group of people
over a period of time. CPI is also used
to calculate the inflation rate and cost of
living. Beside food and drinks, what are
other goods and services that can be
bought by Malaysian households?
10.1.1
Scan this QR code
to watch a video on
Consumer Price Index
(CPI) in Malaysia.
bit.ly/2PvIIMt
271
CHAPTER 10
In general, index number is used to measure
all types of quantitative changes in the field of
industries, agricultures, trades and services.
Besides these, index number also plays an
important role in measuring the magnitude of
the economy such as income, job opportunity,
export, import, price and others.
10.1 Index Numbers
Defining and describing index numbers
I paid RM680 for
this smart phone
last year.
I paid RM748 for
the same smart
phone this year.
Based on the conversation above, what conclusion can be made
regarding the prices of the smart phones in this year and last year?
If you are able to state an increase of 10% in the prices, then you
have made a relation regarding the index number.
In general, index number is a statistical measure to measure
the change of a variable of a particular year as compared to
another year which is considered to be the base year. The base
usually has the value of 100 and index number is 100 times the
ratio with the base. The variables can consist of value of currency,
price, product, earning, quantity, job opportunity and others.
Mathematics Museum
The earliest recorded
calculation of index number
was in 1750.
There are varieties of index numbers with different calculations. For example:
CHAPTER 10
Consumer price index
Market price at the
current year
IHP =
Market price at base year
Air quality index
I –I
I = high low (C – Clow) + Ihigh
Chigh – Clow
I = Air quality index
C = Concentration of pollutants
272
× 100
Death index of road accidents
a
I=
× 10 000
∑vehicles
a = total death at the current year
∑vehicles = total number of registered
vehicles till the current year
Body mass index
BMI =
Weight (kg)
× 100
Height (cm) × Height (cm)
10.1.1
10.1.2
Index Numbers
InQuirY 1
In groups
21st Century Learning
Aim: To determine the percentage change and relate with index numbers
Instruction:
1. Work in 5 groups.
2. Look at the graphical info below regarding the data of total foreign workers according
to the sectors in Malaysia in the years 2013 and 2016.
649
388
268
253
Data of the
194
2016
foreign workers
in Malaysia
Construction Manufacture Agriculture Plantation Services
according to
the sectors
in the years
2013
174
2016 and 2013
269
(in nearest thousand)
432
434
752
[Source: Malaysia Open Source]
3. Every group is required to pick only one sector to analyse.
4. In groups, answer the following questions:
(a) Determine the percentage change in the data for the year 2016 compared to that of
the year 2013 for each sector and interpret the percentage change obtained.
(b) List down the causes for the changes to take place.
(c) State two implications of the entry of foreign workers to our country.
(d) List down suggestions to overcome the negative effects of entry of the foreign
workers in this country.
5. Present your work in a creative way in front of the class.
6. Carry out a question and answer session with the members of other groups.
From the results of Inquiry 1, with the year 2013 as the base year, the index number is the
percentage change of the data of the foreign workers in Malaysia in the year 2016 compared to
the year 2013.
388
× 100%
434
= 89.4%
The above percentage can also be written in index number, I:
388
I=
× 100
434
I = 89.4
10.1.1
10.1.2
CHAPTER 10
Percentage change in data of foreign workers in construction sector =
273
In general, the formula for index number can be written as:
I=
Q1
× 100
Q0
with Q0 = Price/Quantity at the base year
Q1 = Price/Quantity at a particular year
MATHEMATICS
POCKET
Price index or quantity is
the ratio in percentage,
without writing the
percentage symbol.
Example 1
The price of an X-branded watch in the year 2017 and 2018
was RM500 and RM550 respectively. Calculate the index
number of the price of the watch in the year 2018 based on the
year 2017. Interpret the index number obtained.
Solution
Let Q0 = Price in the year 2017
Q1 = Price in the year 2018
Q
Index number, I = 1 × 100
Q0
550
× 100
=
500
= 110
Thus, there is an increase of 10% in the price from the year 2017
to the year 2018.
Example 2
In the year 2017, the number of sports governing bodies registered
with the Sports Commissioner Office is 893. Given that the index
number of the registration of sports governing bodies of the year
2017 based on the year 2010 is 156.39, calculate the number of
sports governing bodies registered in the year 2010.
Smart TIPS
The value of index number
more than 100 means there
will be an increase when
compared to the base year
whereas index number
smaller than 100 means
there will be a decrease or
reduction when compared to
the base year.
Mind Challenge
Can the value of index
number be 100? If so, when
will this arise?
CHAPTER 10
Solution
Let Q0 = The number of registration in the year 2010
Q1 = The number of registration in the year 2017
Q
I = 1 × 100
Q0
156.39 = 893 × 100
Q0
Q0 = 571
Thus, the number of sports governing bodies registered in the
year 2010 is 571.
274
10.1.1
10.1.2
Index Numbers
QR
Example 3
The price index of a bicycle in the year 2018 based on the year
2010 and 2015 was 176 and 110 respectively. Find the price
index of the bicycle in the year 2015 based on the year 2010.
Solution
Q2018
× 100 = 176 …
Q2010
Q2018
× 100 = 110 …
Q2015
Q2015 176
=
1 ÷ 2:
Q2010 110
Q
I = 2015 × 100
Q2010
176
× 100
=
110
= 160
Alternative Method
1
2
Scan this QR code for
other methods to solve
problems involving index
numbers.
I2018/2010
× 100
I2015/2010
176
× 100
110 =
I2015/2010
I2015/2010 = 176 × 100
110
= 160
I2018/2015 =
bit.ly/2Ntoh02
Self Practice 10.1
1. Malaysia Automotive Association (MAA) reported that the total number of registered
commercial vehicles in the year 2015 was 75 376 whereas the total registered number
commercial vehicles in the year 2017 was 61 956. Calculate the index for the number of
registered commercial vehicles in the year 2017 based on the year 2015 and interpret it.
2. The average monthly expenditure of a Malaysian household in the year 2014 was RM3 578.
In the year 2017, average monthly expenditure of a Malaysian household was RM4 033. Find
the average index for the monthly expenditure of a Malaysian household in the year 2017
based on the year 2014 and interpret your findings.
3. The total production of oil palm in Malaysia in the year 2013 was 720 440 105 metric tonnes.
Given the index of the total production of oil palm in the year 2016 based on the year 2013
was 90.23, find the total production of oil palm in the year 2016.
4. The table below shows the price indices for a particular type of drink.
Year 2019
(2011 = 100)
Year 2019
(2013 = 100)
150
225
p
Find the value of p.
Smart TIPS
The year 2013 (2011 = 100)
means price index in the
year 2013 is based on the
year 2011.
5. The production index of industrial sugar in the year 2011 and 2012 based on the year 2010
was 101.4 and 95.8 respectively. Calculate the index production of industrial sugar in the year
2012 based on the year 2011.
10.1.1
10.1.2
275
CHAPTER 10
Year 2013
(2011 = 100)
Problem solving involving index numbers
Example 4
Mathematics
Application
According to the statistics from the Ministry of Natural Resources and
Environment, the total number of visitors who visited Taman Negara
Pahang, Sungai Relau in the year 2016 was 17 721. If National Park
Corporation targeted a 10% increase in the number of visitors for the
year 2018, calculate the expected number of visitors in the year 2020
if the rate of increment of the visitors from the year 2018 to the year
2020 is the same as the rate of increment of the visitors from the year
2016 to the year 2018.
Solution
1 . Understanding the problem
Number of visitors in the year 2016
was 17 721.
◆ The increment of 10 percent from the
year 2016 to the year 2018.
◆ The increment of 10 percent from the
year 2018 to the year 2020.
◆ Find the number of visitors in the year
2020.
◆
2 . Planning the Strategy
Find the number of visitors in the
year 2018 by using the index number
formula.
◆ By using the number of visitors in the
year 2018, the number of visitors in
the year 2020 is calculated using the
index number formula.
◆
3 . Implementing the strategy
The number of visitors in the year 2018
Q
I2018/2016 = 2018 × 100
Q2016
Q
110 = 2018 × 100
17 721
Q2018 = 19 493
The number of visitors in the year 2020
Q
I2020/2018 = 2020 × 100
Q2018
Q
110 = 2020 × 100
19 493
Q2020 = 21 442
Thus, the expected number of visitors in
the year 2020 is 21 442.
4 . Making a conclusion
21 442
× 100 ≈ 110
19 493
19 493
× 100 ≈ 110
◆ The index number in the year 2018 is based on the year 2016,
17 721
CHAPTER 10
◆
The index number in the year 2020 is based on the year 2018,
276
10.1.3
Index Numbers
InQuirY 2
In groups
21st Century Learning
Aim: To study the usage of index number
Instruction:
1. Read the newspaper extract below carefully.
The alarming rate of accidents among people
BANGI: National Institute of Occupational Safety
and Health (NIOSH) expressed their concerns over
the alarming rate of accidents among people in the
country from 66 618 cases in the year 2016 to 69 980
cases in the year 2017.
The chairman of NIOSH, Tan Sri Lee Lam Thye
said, according to the statistics released by Social
Security Organisation (SOCSO), as many as 33 319
cases were recorded in 2017 involving accidents
which occurred when travelling to or returning from
workplaces, the increase was as much as 6.4% from
31 314 cases of accidents recorded in the year 2016.
On the other hand, the rate of occupational hazards
increased as much as 3.84% from 35 304 cases in the
year 2016 to 36 661 cases in the year 2017.
“This increase is particularly worrying as we
celebrated our 61st Independence Day, we are only
independent or free from colonial shackles, but
still not independent from the aspect of attitude
especially on the road,” he said in the media
conference after officiating celebration of the 61st
Independence Day 2018 at NIOSH headquarters.
Translated from Berita Harian
(Source: https://www.bharian.com.my/berita/
nasional/2018/08/468225/kadar-kemalangan-dikalangan-rakyat-membimbangkan)
2. Carry out a brainstorming session among group members and answer the following
questions:
(a) Make a conjecture regarding the occupational hazard index in the year 2017
compared to the year 2016.
(b) What are the effects if the rate of occupational hazards continue to increase?
(c) What are the causes of the increase of occupational hazards in our country?
(d) Suggest ways to reduce the rate of occupational hazards in our country.
3. Prepare a graphical folio to answer the above questions.
4. Display your group result to the class.
Self Practice 10.2
1. The table shows the price index of groceries in the year 2015 and 2020 based on the year
2010.
Price index in the year
2015
2020
Groceries
125
140
Find the price index of these groceries in the year 2020 based on the year 2015.
2. The premium insurance payment of a company in the year 2016 increased as much as 5
percent compared to the year 2011. In the year 2018, the premium increased again as much as
10 percent compared to 2011. Find the premium insurance index in the year 2018 compared
to the year 2016.
10.1.3
277
CHAPTER 10
Item
Intensive Practice 10.1
Scan the QR code or visit bit.ly/2IxHTyD for the quiz
1. In January 2017, the average temperature in town P was 25.3°C whereas the average
temperature in February 2017 was 27.4°C. Find the average temperature index in February
with January as the base time and interpret the index number obtained.
2. Given the price index of a certain item in the year 2016 based on the year 2015 was 130 and
the price index in the year 2016 based on the year 2012 was 120. Find the price index of this
item in the year 2015 based on the year 2012 and interpret it.
3. The table below shows the prices and the price indices of three ingredients P, Q and R which
are used in preparing a type of biscuit.
Cost (RM/kg)
Year 2015
Year 2019
Price index in the year 2019
based on the year 2015
P
x
0.40
80
Q
2.00
y
140
R
0.80
1.00
z
Material
Find the values of x, y and z.
4. The table below shows the retail prices of a chicken in January for the year 2015 until 2018.
Year
Price (RM/kg)
Price index
2015
5.80
p
2016
7.65
q
2017
7.80
r
2018
7.30
s
Using the year 2015 as the base year, find the value of p, q, r and s.
5. The diagram below shows the price index for a type of food in 2015 and 2018 based on the
year 2010.
CHAPTER 10
Item
Food
Price index
Year 2015
Year 2018
110
118
Find the price index of the food in the year 2018 based on the year 2015.
278
Index Numbers
10.2 Composite Index
Determining and interpreting composite index
InQuirY 3
In pairs
21st Century Learning
Aim: Determine composite index
Instruction:
1. The table below shows the price indices and percentages of four ingredients used in
preparing semperit in the year 2019 based on the year 2018.
Ingredient
Price index
Percentage (%)
Margarine
120
30
Sugar
127
15
Wheat flour
108
50
Egg
107
5
2. Calculate the average price index for all the four ingredients and make a conclusion
about the average value.
3. What is the role played by the value of percentages in the calculation of the average price
index? If the percentages of these four ingredients are the same, what can you interpret
about the average price index?
4. Present your findings in front of the class and carry out a question and answer session
with other pairs.
From Inquiry 3, the average price index is obtained as follow:
(120 × 30) + (127 × 15) + (108 × 50) + (107 × 5)
100
= 114.4
Average price index =
The average price index indicates that there is an increase in the price of the raw ingredients
in the year 2019 compared to the year 2018. The value of the percentages represents the
importance of the usage of the raw materials in the preparation of semperit.
The value of this average price index is known as the composite price index (I) which
means the combination of a few indices as a statistical measure for overseeing the market or
sector performance from time to time involving the importance of each item. The importance
is known as the weightage (w). The value of weightage can be in the form of numbers, ratios,
percentage, readings on the bar chart or pie chart and others.
If I1, I2, I3, ..., In are the price indices for n items respectively with weightages w1, w2, w3,
..., wn, then the composite index can be calculated using the following formula:
(I1w1 + I2w2 + I3w3 +… + Inwn)
w1 + w2 + w3 +… + wn
∑Iiwi
I =
∑wi
with Ii = index numbers and wi = weightages
10.2.1
CHAPTER 10
I =
279
Example 5
Price index of one kilogram of three types of fruits sold in a stall
in the year 2018 based on the year 2010 was 175, 120 and 160
respectively. Find the composite index of these fruits in the year
2018 based on the year 2010.
MATHEMATICS
POCKET
Composite index without
weightage given is
calculated by assuming the
value of the weightages
are the same for each index
number.
Solution
∑Iiwi
∑wi
175(1) + 120(1) + 160(1)
I =
3
= 151.67
Composite index, I =
the weightage
for each type
of fruits is 1
Mind Challenge
What is the difference
between composite index
with and without weightages
given? Explain the
importance of weightage in
the calculation of composite
index.
Example 6
The table below shows the expenditure of utility index of a factory in the year 2017 based on
the year 2011. The pie chart shows the percentages of the usage in a month.
Utility
Expenditure index
Water
135
Electricity
140
Fuel
125
Fuel
25%
Water
17%
Electric
58%
Find the composite index of the expenditure of utility in the
year 2017 based on the year 2011.
Solution
∑Iiwi
∑wi
135(17) + 140(58) + 125(25)
=
17 + 58 + 25
13 540
=
100
= 135.4
CHAPTER 10
Composite index, I =
280
10.2.1
Index Numbers
Self Practice 10.3
1. The price index of the traditional kuih such as nekbat, nagasari and serabai in the year 2020
based on the year 2015 is 105, 112 and 98 respectively. Find the composite index for the
three types of traditional kuih in the year 2020 based on the year 2015 and interpret the value
obtained.
2. The bar chart shows the credit points of three subjects
in a college. Given the student entry index following
the subjects Chemistry, Economics and Science in the
year 2019 based on the year 2015 was 136, m and 108
respectively. Find the value of m if the composite index
for the three subjects in the year 2019 based on the year
2015 was 120.
Weightage
3
2
1
0 Chemistry Economy Science
Subject
Solving problem involving index numbers and composite numbers
The concepts of index numbers and composite numbers studied this far are used widely in
various fields for recognising and monitoring the trend in prices, production, job opportunities,
inflation and others.
Example 7
The table below shows the cost price of three main materials in making non-rusting steel by a
company.
Material
Iron
Chromium
Nickel
Price in the year 2010
(RM per metric tons)
2 025
8 431
62 235
Price in the year 2018
(RM per metric tons)
3 424
9 512
50 916
Percentage
(%)
72
18
10
(a) Calculate the price index for iron, chromium and nickel in the year 2018 based on the year
2010.
(b) Calculate the composite index for the cost of the production of the non-rusting steel in the
year 2018 based on the year 2010. Interpret your findings.
(c) Determine the cost of production of the non-rusting steel in the year 2018 if the cost in the
year 2010 was RM65 million.
Solution
Q2018
Q
Q
× 100 IChromium = 2018 × 100 INickel = 2018 × 100
Q2010
Q2010
Q2010
3 424
9 512
50 916
× 100
=
× 100
=
× 100
=
2 025
8 431
62 235
= 169.09 = 112.82 = 81.81
Thus, the price index for iron, chromium and nickel in the year 2018 based on the year
2010 was 169.09, 112.82 and 81.8 respectively.
10.2.1
10.2.2
281
CHAPTER 10
(a) IIron =
(b) Construct a table to determine ∑wi and ∑Iiwi.
Material
Ii
wi
Iiwi
Iron
169.09
72
12 174.48
Chromium
112.82
18
2 030.76
Nickel
81.81
10
818.10
∑wi = 100
∑Iiwi = 15 023.34
I = ∑Iiwi
wi
15
023.34
=
100
= 150.23
There was an increase of 50.23% in the production cost of the non-rusting steel in the year
2018 based on the year 2010.
Q
BRAINSTORMING
I = 2018 × 100
(c)
Q2010
Did the decreasing price of
Q
the nickel in 2018 affect the
150.23 = 2018 × 100
65
total production cost
Q2018 = 97.65
of the production of the
Thus, the production cost of the non-rusting steel in the
year 2018 was RM97.65 million.
non-rusting steel? Discuss.
Self Practice 10.4
1. The table below shows the price of four materials, A, B, C and D used in making roof tiles in
the year 2016 and 2010.
Material
CHAPTER 10
A
B
C
D
2010
1.40
1.50
1.60
4.50
Cost (RM)
2016
2.10
1.56
1.92
5.58
Weightage (%)
10
20
40
30
(a) Calculate the price index for each material in the year 2016 based on the year 2010.
(b) Calculate the composite index for all the materials in the year 2016 based on the year
2010. Interpret your findings.
(c) Determine the price of the roof tile in the year 2010 if the price was RM2.65 in the year
2016.
282
10.2.2
Index Numbers
2. The table below shows the price of five materials used in making a souvenir in the year 2013
and 2019.
Material
P
Q
R
S
T
Price in the
year 2013
(RM)
5.00
20.00
8.00
16.00
10.00
Price in the
year 2019
(RM)
6.00
23.00
12.00
18.00
13.00
Price index
(2013 = 100)
Weightage
(%)
120
a
b
c
130
8
12
20
27
d
(a) Calculate the value of a, b, c and d.
(b) Calculate the composite index for the souvenir in the year 2019 based on the year 2013.
Interpret your findings.
(c) Determine the price of the souvenir in the year 2019 if the price in the year 2013 was
RM35.
(d) Calculate the price index of the souvenir in the year 2021 if the total cost of the materials
used is expected to increase by 10% in the year 2021.
Intensive Practice 10.2
Scan the QR code or visit bit.ly/2Ox3nPM for the quiz
1. The admission of students in a school according to the Science stream and the Arts stream
follows the ratio 60 : 40. Given that the admission index of students according to the Science
stream and the Arts stream in the year 2019 based on the year 2015 was 120 and 130
respectively. Find the composite index for the admission of students in the school in the year
based on the year 2019 and 2015.
2. Myra Company has three small subsidiaries in three districts in Selangor. The table below
shows the change in productivity and the number of workers in the three subsidiaries in the
year based on the year 2018 and 2010.
District
Kuala Langat
Gombak
Shah Alam
Change in productivity from the
year 2010 to 2018
Increased 10%
No change
Decreased 20%
Number of
workers
3
2
5
3. The subject evaluation in a college follows the Paper 1, Paper 2 and Course Work format. The
allocation for Course Work is 20% of the total marks of the subject whereas the marks for Paper
1 and Paper 2 is 80% of the total marks and both of the papers are important in the calculation
of the final marks. Kalaivathy obtains 85, 72 and 68 marks for Paper 1, Paper 2 and Course
Work respectively. Calculate the final marks obtained by Kalaivathy in the subject.
10.2.2
283
CHAPTER 10
Find the composite index for the productivity of the three subsidiaries. Give your opinion
regarding the productivity of Myra Company based on the value you have obtained.
4. The table below shows the price index and change in the price indices for four main materials
in the production of a facial wash.
Price index in the
year 2021 based on
the year 2019
Change in the price
index from the year
2021 to the year 2023
B
140
Reduced by 10%
C
m
Material
150
A
No change
C
30%
No change
115
D
B
40%
A
10%
D
20%
Increased by 20%
(a) Find the value of m if the composite index in the production of facial wash in the year
2021 based on the year 2019 is 133.
(b) Calculate the composite index in the production of the facial wash in the year 2023 based
on the year 2019.
(c) Calculate the production cost of the facial wash in the year 2023 if the cost corresponds
to the year 2019 is RM19.50.
Summary OF Chapter 10
i��
�d
u�
��
f�
i�
INDEX
NUMBERS
Composite index
With weightage
��a
�m
�
o��e
i��
�
��e
�o
Without weightage
Index number
f�
m�
�
�a
I=
Q1
Q0
100
� ��
�� �
��e��i�
∑ Iiwi
,
∑wi
wi = weightage
I
=
CHAPTER 10
WRITE YOUR JOURNAL
Based on your comprehension throughout this chapter, what do you understand about index
number? In your opinion, what is the best way to determine the most suitable base year in
calculating the index number of goods or services? What is weightage? What are the factors
that affect the relative importance of an item?
284
Index Numbers
MASTERY PRACTICE
1. The table below shows the price per kg of four types of goods, A, B, C, and D, in the year
2017 and 2019, price index in the year 2019 was based on the year 2017 and their weightages
respectively. PL3
Good
A
B
C
D
Price in the
year 2017
(RM/kg)
2.00
0.80
1.10
x
Price in the
year 2019
(RM/kg)
2.20
y
1.10
1.20
Price index in the
year 2019
(2017 = 100)
z
125
100
120
Weightage
4
1
2
3
(a) Find the value of x, y and z.
(b) Calculate the composite index of the goods in the year 2019 based on the year 2017.
2. The table below shows the price index of two materials A and B used in production of a type
of household decorations. PL3
Material
A
B
Price index in the year 2018
based on the year 2016
110
n
Price index in the year 2020
based on the year 2016
m
110
Given the price of material B increased by 22% in the year 2018 from the year 2016. The
price of material A in the year 2016 was RM5.00 and the year 2020, RM6.05. Find the value
of m and n.
3. The table below shows the information related to four materials, A, B, C and D used in making a
toy. The percentage of usage of material B is not shown. PL4
A
B
C
D
Change in price index from the
year 2015 to the year 2018
Reduced by 10%
Increased 60%
Increased 20%
Increased 40%
Percentage usage (%)
50
10
10
The production cost of the toy was RM41 650 in the year 2018.
(a) If the cost of material C in the year 2015 was RM7.60, find the cost in the year 2018.
(b) Calculate the corresponding production cost in the year 2015.
(c) The production cost is expected to increase by 60% from the year 2018 to the year 2020.
Calculate the percentage change in the production cost from the year 2015 to the year 2020.
285
CHAPTER 10
Material
4. The rubber production in Malaysia is 1.126 million tonnes in 2005, x million tonnes in 2010
and 0.722 million tonnes in 2015. Calculate PL3
(a) the index number for the rubber production in the year 2015 based on the year 2005,
(b) the value of x, given that the index number for rubber production in the year 2010 based
on the year 2005 is 83,
(c) the index for the rubber production in the year 2020 based on the year 2005 if the index
for rubber production in the year 2020 based on the year 2010 is 105.
5. The table below shows the price for an item in 2000 and 2015.
Year
2000
2015
PL4
Price
RM8
RM10
(a) If the rate of price increase from 2015 to 2020 is twice the rate of price increase from
2000 to 2015, determine the price of that item in 2020.
(b) Calculate the price index in the year 2020 based on the year 2000.
6. The table below shows the price index and weightage for four types of materials in the year
2020 based on the year 2019. PL4
Materials
Price index
Weightage
P
107
2
Q
118
x
R
94
1
S
105
2x
(a) The composite index for the materials in the year 2020 based on the year 2019 is 108.
Determine the value of x.
(b) The price index for material P rises by 20% and the price index for material S drops by
10% in 2020 until 2021. The price index for other materials do not change. Determine the
composite index for the materials in the year 2021 based on the year 2019.
7. The table below shows the sales index for an encyclopedia in 2015 and 2017 with the year
2000 as the base year. PL4
Year
Sales index
2015
109
2017
145
Determine the sales index for the encyclopedia in the year 2017 based on the year 2015.
8. The table below shows the price index for three cameras.
CHAPTER 10
Camera
Year
J
K
L
2013
(2011 = 100)
165
q
150
Determine the value of p, q and r.
286
PL4
2019
(2011 = 100)
231
156
r
2019
(2013 = 100)
p
120
170
Index Numbers
9. The following shows the number of visitors who visited Pulau Langkawi in 2010 and 2017.
PL5
2010
2017
2.45 million
3.68 million
(a) Determine the number of visitors in 2020 if the rate of increase for the number of visitors
in 2017 to 2020 is twice the rate of increase from 2010 to 2017.
(b) Calculate the index for number of visitors in the year 2020 based on the year 2017. State
your interpretation based on the index number obtained.
10. The table below shows the price and weightage for three materials P, Q and R in the year
2018 based on the year 2016. PL4
Material
Price index
Weightage
P
80
x
Q
130
y
R
140
z
Given the composite index for material P and Q in the year 2018 based on the year 2016 is
120 whereas the composite index for materials P and R is 125. Determine the ratio x : y : z.
11. The price index for a safety helmet in the year 2014 based on the year 2010 was 80 and the
price index in the year 2018 based on the year 2014 was 110. Given the price for the safety
helmet in the year 2018 was RM166. PL5
(a) Calculate the price of the safety helmet in 2010 and 2014.
(b) Determine the percentage of decrease in price for safety helmets in 2010 compared to its
price in 2018.
12. The price for service charge in an agency in 2018 was RM1.50. If the price increased by 15%
in 2019, calculate PL5
(a) the price index for service charge in 2019 with 2018 as the base year,
(b) the price for the service charge in 2020 if the rate of price increase in 2019 to 2020 is the
same as the rate of price increase for 2018 to 2019.
Exploring
MATHEMATICS
1. Prepare a monthly expenditure of your family according to each of the following
categories for the span of 3 months.
(d) Transportation
(a) Food and beverages
(b) Clothing and shoes
(e) Medication
(c) Water and electricity bills
(f) Education
3. Determine the composite index for expenditures on the 2nd and 3rd month based on the 1st
month. What conclusion can you make based on the composite index value obtained?
4. Explain ways to spend wisely.
5. Discuss in groups and make an interesting graphic folio.
287
CHAPTER 10
2. Explain the weightage based on the relative money spent by your family.
Open the complete answer file in the QR Code on page vii to get the full solutions
Chapter 1 FUNCTIONS
Self Practice 1.1
1. (a) Function because every object has only one
image even though element 7 has no object.
(b) Function because every object has only one
image even though element 4 has two objects.
(c) Not a function because object r has two
images, 8 and 10.
(b) Not a function
2. (a) Function
(c) Function
1
(b) h : x → |x|
3. (a) h : x → , x ≠ 0
x
(c) h : x → x3
Self Practice 1.2
1. (a) Domain = {–2, –1, 0, 2, 4}
Codomain = {1, 3, 4, 5}
Range = {1, 3, 4, 5}
(b) Domain = {j, k, l, m}
Codomain = {2, 3, 6, 7, 10}
Range = {3, 7}
(c) Domain of f is –3  x  5
Codomain of f is 2  f(x)  6
Range of f is 2  f(x)  6
2. (a) f(x)
f(x) = |x + 1|
5
Range of f is 0  f(x)  5.
–2 –1 0
(b) f(x)
1
x
4
f(x) = |4 – 2x|
–2 0
2
f(x)
Range of f is 0  f(x)  8.
x
4
(c)
9
5
3
–2 0
288
( 12 ) = –9
1. (a) g(–5) = 2, g(–2) = 1, g
1
(b) b = – , b = 3
2
2. (a) k = 4
(c) k = 3
( 12 ) = 5 (b)
3. (a) f(–2) = 11, f –
(c)
1
<x<1
2
(b) k = 3
4. 2, 6
5. (a) m = – 4, c = 15
(b) 7
1
,1
2
1
(d) x < – , x > 2
2
(c) 3
Intensive Practice 1.1
1. (a) and (c) because every object has only one
image.
2. (a) Not a function
(b) Not a function
(c) Function
3. (a) Function because every object has only one
image.
(b) Domain = {–7, –6, 6, 7}
Range = {36, 49}
(c) f : x → x2
4. (a) t = 6
(b) 0  f(x)  6
(c) 0  x  4
5. (a) (i) 80 meter
(ii) 72 meter
(iii) 45 meter
(b) 3 seconds
Self Practice 1.4
8
4
Self Practice 1.3
f(x)=|2x – 5|
5–
2
4
x
Range of f is 0  f(x)  9.
1. (a)
2. (a)
(b)
f(x) = 3x
(b) gf(x) = 2x – 7
fg : x → 9 – 3x, gf : x → 3 – 3x
f 2 : x → 9x, g2 : x → x
fg : x → 4 + 2x2, gf : x → 4x2 + 16x + 16
f 2 : x → 4x + 12, g2 : x → x4
6
6
, x ≠ –4
(c) fg : x → + 4, x ≠ 0, gf : x →
x
x+4
f 2 : x → x + 8, g2 : x → x
6 – 5x
1
, x ≠ 1, gf : x →
,x≠6
(d) fg : x →
x–1
x–6
x–1
f 2 : x → x – 10, g2 : x →
,x≠2
2–x
2
2
3. fg(x) = 3x + 22, gf(x) = 9x + 24x + 22
(a) x = 1, x = 2
(b) x = 0, x = –4
4. a = –2, b = 9 or a = 2, b = –3
5. h = –k
Self Practice 1.5
1. (a) fg(3) = 4
( 12 ) = –1
(c) f 2(4) = 3, g2
1
2
2. (a) x = 2
(c) x = 2
(d) f 2(–1) = 5, g2(–1) = –
( 15 )= 9
(b) gf –
(b) x = 2, x = –2
(d) x = 1
Self Practice 1.6
1. (a) g : x → 2x2 – 4x + 10
(b) g : x → x + 2
2. (a) g : x → x2 – 4x
(b) g : x → 2x – 3
2
(b) x = 24
3. (a) g : x → , x ≠ 0
x
4. (a) f (x) = 3x – 7
(b) gf(2) = –3
Self Practice 1.7
x
1
,x≠–
2x + 1
2
x
1
3
,x≠–
f (x) =
3x + 1
3
x
1
4
,x≠–
f (x) =
4x + 1
4
x
1
20
,x≠–
f (x) =
20x + 1
20
x
1
,x≠–
f 23(x) =
23x + 1
23
f 2(x) = x
1
f 3(x) = , x ≠ 0
x
f 4(x) = x
f 40(2) = 2
1
f 43(2) =
2
16 6
7
πt
(b) 113 π m2
Ar(t) =
9
9
v
(i) v(t) = 200 + 100t (ii) h = 2
πr
2+t
(iii) hv(t) =
4π
1.75 cm
r(t) = 3t
Ar(t) is the area of water ripple, in cm2, as
function of time, t in seconds
8100 π cm2
1. (a) f 2(x) =
(b)
2. (a)
(b)
3. (a)
4. (a)
(b)
5. (a)
(b)
(c)
Intensive Practice 1.2
x–1
, x ≠ –1
x+1
2x – 1
gf(x) =
,x≠0
2x
1
(b) fg(2) =
3
1
=2
gf –
2
1
(c) x =
3
1. (a) fg(x) =
( )
2. (a) h = 3, k = –1
(b)
5
6
3. a = 2, b = 3
4. (a) h = 2, k = −3
− 3x2 + 18x − 19
(b) gf(x) =
,x≠3
(x – 3)2
5. (a) a = 3, b = 1
(b) f 4(x) = 81x + 40
6. (a) A(x) = x2, V(A) = 10A
7. (a) g(x) = 2x2 – 3x – 13
(b) g(x) = x2 – 12x + 40
(c) g(x) = 14 – x
x–1
(b) f (x) → 9x2 – 3x + 4
8. (a) g : x →
3
9. (a) p = 2, q = –1
(b) f 4(x) = 16x – 15
(c) f n(x) = 2nx + 1 – 2n
10. CN(t) = 15 000 + 800 000t – 40 000t2
Self Practice 1.8
1. (a) f(4) = –5
(c) f –1(2) = –2
1
2. (a) g(12) = –
2
(c) h(–1) = 3
(b) f –1(–1) = 6
(d) f –1(–5) = 4
3
(b) g–1(4) =
4
(d) h–1(9) = 1
Self Practice 1.9
1. (a) Has an inverse
(b) Does not have an inverse
(c) Does not have an inverse
(d) Has an inverse
(e) Does not have an inverse
(f) Does not have an inverse
(g) Has an inverse
2. (a) Inverse function
(b) Inverse function
(c) Not an inverse function
(d) Not an inverse function
3.
f(x)
8
f
2
1
−1 0
−11 2
f −1
8
x
The domain of function f –1 is –1  x  8 and the
range is –1  f –1(x)  2.
y
4. (a)
7
h
h −1
3
−2 0
−2
3
7
x
(b) The domain of function h–1 is –2  x  7
(c) x = 2
1
(b) Qʹ(–3, 1)
5. (a) Pʹ , –2
2
(c) Rʹ(5, 4)
(d) Sʹ(–8, – 6)
( )
289
6. (a)y
(b) a = 1, b = 4
8. (a)
r
y = f −1( x ) Bʹ(10, 3)
0
x
Aʹ(2, − 1)
f –1
Self Practice 1.10
x+5
2
4+x
–1
(c) f : x →
,x≠0
x
8x
+
9
(e) f –1 : x →
,x≠1
x–1
1
2. (a) f –1(4) =
3
5
1
3. a = – , b =
2
8
x–7
4. (a) f : x →
6
3x
(c) f : x →
,x≠3
x–3
1. (a) f –1 : x →
5. (a) k = 2
3
,x≠0
x
6x
(d) f –1 : x →
,x≠5
x–5
x
–
3
(f) f –1 : x →
,x≠1
2x – 2
3
(b) x = – , x = 1
2
(b) f –1 : x →
f (2) = 5
gf (2) = 8
g–1(8) = 5
Yes
No
y
()
1
(b) g
= –6
2
(b)
(d)
(f)
(b)
f
(b)
(4, 2)
y
4
f
The domain of function
f –1 is 0  x  4.
f −1
0
2
y
4
x
(4, 8)
f
(8, 4)
f −1
0
4. (a) h = 5
(c) m = –9
3x – 2
,x≠0
5. (a) h(x) =
x
6. x = –5, x = 2
20
7. (a) f –1(x) = 220 – x
17
290
The domain of function
f –1 is 0  x  2.
x
0
2
(c)
g(5) = 8
f –1(5) = 2
f –1g–1(8) = 2
Yes
(2, 4)
f −1
Mastery Practice
1. (a) (i) 1
(ii) 6, 8, 9
(b) Yes because every object only has one image.
(c) Domain = {2, 6, 7, 8, 9}
Codomain = {1, 4, 5}
Range = {1, 4}
2. (a) m = 35
(b) h : x → x2 − 1
3. Function but not a one-to-one function.
y
4. (a)
4 f ( x ) = x − 3
(b) f : x → 2 – 5x
Intensive Practice 1.3
1. (a)
(c)
(e)
2. (a)
(c)
3. (a)
(b) 0.49 cm
V = 4– πr3
3
f
The domain of function
f –1 is 0  x  8.
−1 0
3
The range of function f is 0  f(x)  4.
(b) 1  x  5
y
(c)
y = 2x−3
f ( x ) = x − 3
4
x = 2
5. (a) h = 7, k = 6
6. m = 3, c = –13
7. (a) (i)
(b) 43
f –1(x) =
x+2
3
(ii)
g(x) = 2x + 5
(b) x = –8
8. (a) k = 1
(b) m = 2, n = 1
(c) f 2(x) = x
(d) f –1(2) = 3
9. (a) (i) Continuous functions (ii) – 4  f(x)  4
(b) Does not have an inverse function
1
10. (a) Condition x  0.
(b) f –1(x) = x, f –1(x) = x4
11. A graph does not have to cross the line y = x if the
graph of a function and its inverse intersect. Both of
these graphs might intersect at another line.
8 + 5x
,x≠1
12. (a) (i) f –1(x) =
x–1
–3 – 4x
(ii) f –1(x) =
,x≠2
x–2
(b) f = f−1 if a = −d
13. (a) (i) (ii)
x
y=x
f –1
2 ( 0, 2 )
(b) f –1(3) = 14
(b) 173.4
x
7
3
− 10
−3
3
(b) x = 2
x
7
f
1
–1
0
–1
( 2, 0 )
1
The range of f is
–1  f(x)  3 and the
domain of f –1 is
–1  x  3.
3
(b) Range of f = domain of f –1 and domain of
f = range of f –1.
(i)
(ii)
Yes
Yes, any points of (b, a) on the graph of f –1
are the reflection points of (a, b) on the
graph of f at the line y = x.
2
100 – p + 600
14. (a) C = !w
25
(b) RM600.64
15.
14
4 x 0.5
h(x) = 2π –––
10 12
10
x + 4 0.5
g(x) = 2π –––––
10 8
x 0.5
f(x) = 2π –––
10 6
4
2
–5
0
5
10
15
The period of pendulum T depends on the length of
the pendulum, l. If the length increases, the period of
oscillation of the pendulum also increases.
Chapter 2 QUADRATIC FUNCTIONS
6.
7.
8.
9.
10.
11.
12.
13.
–5.606, 1.606
−7.243, 1.243
0.134, 1.866
−1.317, 5.317
0.131, 2.535
−0.449, 4.449
8 cm, 6 cm
(b)
(d)
(f)
(b)
(d)
(f)
(b)
−1.193, 4.193
0.634, 2.366
−0.712, 4.212
−1.366, 0.366
−0.425, 1.175
0.275, 2.725
8 cm × 5 cm
Self Practice 2.2
1. (a) x – 8x + 12 = 0
(c) x2 + 11x + 28 = 0
2. p = 2, q = −9
3. (a) 5x2 – 30x + 31 = 0
(c) 5x2 – 14 = 0
4. (a) x2 – 5x – 2 = 0
(c) 4x2 – 29x + 1 = 0
5. 8x2 + 36x – 27 = 0
(b) x2 – 3x – 4 = 0
(d) 5x2 + 24x – 5 = 0
(b)
(d)
(b)
(d)
x2 – 10x – 45 = 0
15x2 – 10x – 3 = 0
2x2 – 5x – 1 = 0
2x2 + 29x + 2 = 0
Self Practice 2.3
1. (a) –2 < x < 2
(c) –2  x  6
(e) –3 < x < 1
2. x  –2 or x  8
(b) 2 < x < 8
(d) x  –1 or x  3
2
<x<4
(f)
3
m = 12; 2, 6
2 and 4; k = 8
−12, 12
h = 2, k = −5
64 – 9d2
c=
4
1
(a) x  − or x  1
2
(c) – 4 < x < 4
(a) m = −1, n = 12
a = 3, b = −10
1. (a)
(b)
(c)
(d)
(e)
(f)
(b) 1  x  4
(b) m = −20, n = 6
12; two different real roots
0; two equal real roots
–104; no real roots
109; two different real roots
0; two equal real roots
49; two different real roots
Self Practice 2.5
1. (a) p = –
2.
3.
4.
5.
3
or p = 3
4
(b) p > –
3
4
3
4
k < –2 or k > 6, k = –2 or k = 6
(a) h = −4, k = −12
(b) c < –16
5
k= h
4
5:4
(c) p <
Intensive Practice 2.2
1. (a) Two equal real roots
(b) Two different real roots
(c) No real roots
2. (a) k = −4 or k = 8
(b) k = −
3. (a) r < −3 or r > 5
(b) r <
4. (a) p <
4
5
1
4
(b) p < −
1
8
1
24
5
5
or k =
(b) x = –3
3
2
6. m = 2n – 4
7. (a) b = 8, c = 12
(b) –6, –2
8. (a) c1 = 4, c2 = 5
(b) The equation does not have two real roots
5. (a) k = −
Self Practice 2.6
1. (a) (i)
Intensive Practice 2.1
1. 0.059, 5.607
2. (a) x2 – 12x + 11 = 0
(b) 12, 11
3. (a) 19x2 – 4x – 1 = 0
(b) 7x2 + 160x + 175 = 0
(c) x2 + 12x + 13 = 0
4. k = −14
(b) r = −3
5. (a) r = 1
1
16
Self Practice 2.4
Self Practice 2.1
1. (a)
(c)
(e)
2. (a)
(c)
(e)
3. (a)
4. 3
(c) r = −2 or r =
The width of the graph decreases,
y-intercept does not change.
y
6
–2
0
y = –x2 + x + 6
3
x
291
(ii)
The width of the graph increases,
y-intercept does not change.
Self Practice 2.8
y
6
y = –x2 + x + 6
x
3
–2 0
(b) The vertex is on the left side of y-axis. All points
are changed except for y-intercept. The shape of
the graph does not change.
y
6
y = –x2 + x + 6
–2 0
3
x
(c) The graph moves 8 units downwards. The shape
of the graph does not change.
y
6
y = –x2 + x + 6
–2 –20
3
1. a = 2, p = 1, q = 5
2. (a) f(x) = x2 – 4x + 3,
f(x) = (x – 1)(x – 3)
(b) f(x) = – 4x2 + 4x + 8,
f(x) = – 4(x + 1)(x – 2)
(c) f(x) = 2x2 + 4x – 16,
f(x) = 2(x + 4)(x – 2)
1
3. The vertex is (–4, –5), f(x) = – x2 – 4x – 13
2
4. (a) a = −1, h = 2, k = 16
(b) f(x) = –x2 – 4x + 12
f(x) = – (x + 6)(x – 2)
1 2 25
–
5. (a) f(x) = x –
2
4
2
(b) f(x) = – (x + 1) + 5
1 2 49
+
(c) f(x) = –2 x +
4
8
2
1
28
–
(d) f(x) = 3 x –
3
3
(e) f(x) = – (x – 2)2 + 16
(f) f(x) = 2(x + 1)2 – 18
( )
( )
( )
Self Practice 2.9
1. (a) The maximum point is (2, 4) and the equation of
the axis of symmetry is x = 2.
(b) (i) When a changes from –3 to –10, the
width of the graph decreases. The axis of
symmetry x = 2 and the maximum value 4
does not change.
x
y
4
f(x)= –3(x – 2)2 + 4
Self Practice 2.7
1. (a) The quadratic function has two
equal real roots. The graph is a
parabola that passes through the
maximum point and touches the
x-axis at a point.
(b) The quadratic function has
two different real roots.
The graph is a parabola that
passed through the minimum
point and intersects the x-axis
at two points.
(c) The quadratic function
does not have real roots.
The graph is a parabola that
passes through the minimum
point and above the x-axis.
2. (a) −1, 2
4. (a) r < –
292
(b) q > –
2
3
(b) r >
0
x
4
3
10
3
x
2
(ii) When h changes from 2 to 5, the graph
with the same shape moves horizontally
3 units to the right. The equation of the
axis of symmetry becomes x = 5 and the
maximum value does not change which is
4.
y
f(x)= –3(x – 2)2 + 4
4
x
(b) 1, 5
3. (a) q < 7
x
0
2
5
x
(iii) When k changes from 4 to –2, the graph
with the same shape moves vertically 6
units downwards. The maximum value
becomes –2 and the axis of symmetry does
not change.
y
f(x)= –3(x – 2)2 + 4
(d)
y
(2, 2)
4
01
x
3
f ( x ) = − 2 (x − 1) (x − 3)
0
x
2
−6
–2
(e)
y
( − 4, 1)
2. (a) h = 3, k = −3, p = 3
(c) −1
(b) x = 5
3. (a) The graph moves 6 units to the right and the
width of the graph increases. The equation of
the axis of symmetry becomes x = 6 and the
minimum value does not change, which is 0.
(b) The graph moves 1 unit to the right and 5 units
upwards and the width of the graph decreases.
The equation of the axis of symmetry becomes
x = 1 and the minimum value becomes 5.
(c) The graph moves 1 unit to the left and 4 units
downwards and the width of the graph increases.
The equation of the axis of symmetry becomes
x = –1 and the minimum value becomes −4.
− 15
(f)
y
f ( x ) = 2 (x + 1) (x − 3)
−1 0
−6
(1, − 8)
(g)
y
y
f ( x ) = (x − 1)2 − 4
−3
(b)
x
3
(2, 9)
f (x) = −x 2 + 4 x + 5
−1 0
(h)
y
f (x) = 2 x 2 + 3x − 2
y
6
−2
f ( x ) = 2 (x + 2) − 2
(i)
x
−3
−1 0
( − 2, − 2)
y
(
0
3, – 31
_
–_
4
8
y
)
(2, 16)
12
5
f ( x ) = 9 − (x − 2)2
5
x
x
1
_
2
−2
(2, 9)
−1 0
x
5
(1, – 4)
2
(c)
x
3
5
−1 0
x
f ( x ) = − (x + 3) (x + 5)
Self Practice 2.10
1. (a)
0
−5 −3
−2 0
f ( x ) = − x 2 + 4 x + 12
6
x
Self Practice 2.11
1. (a) 4 m
(c) 7.2 m
2. (a) 15 m
3. 4 m, 1 m
4. (a) 200 m
(b) 0.8 second
(d) 0 < t < 2
(b) 31.62 m
(b) 50 meter
293
Intensive Practice 2.3
1. (a) k = −1 or k = 4
2. 5
3. (a) (x + 3)2 – 9 + n
(c)
f ( x ) = (x + 3)2 − 5
Mastery Practice
(b) k > –
(b) 4
y
−3
7
3
4
x
0
−5
4. −6 < r < −2, r = –6 or r = –2
5. (a) The width of the graph decreases. The axis
of symmetry and the minimum value do not
change.
(b) The graph with the same shape moves
horizontally 3 units to the right. The equation
of the axis of symmetry becomes x = 4 and the
minimum value does not change, which is 2.
(c) The graph with the same shape moves vertically
3 units upwards. The minimum value becomes 5
and the axis of symmetry does no change, which
is x = 1.
6. (a)
h(t)
t=3
18
(b)
t
3
0
r(t)
36
1. −0.816, 3.066
2. (a) x2 – 8x + 13 = 0
(b) 8, 13
(c) Two real and different roots
(b) k < −8, k > 4
3. (a) k = −8, 4
(c) k  –8, k  4
4. (a) p = 2
(b) p = −1
5. h : k = 7 : 6 ; x = 1
6. x < 2 or x > 5, 0  x  7; 0  x < 2 or 5 < x  7
(b) p = −5, q = −12
7. (a) 3, 7
(c) x = 5
(d) 3 < x < 7
8. (a) b = −8, c = 12
(b) (4, −4)
(c) 2 < x < 6
(d) 4
9. 9 km/h
10. 67.229 units
(b) 20 units
11. (a) 20 units
1
12. (a) y = – (x – 3)2 + 2.5
18
(b) 9.708 m
Chapter 3 Systems of Equations
Self Practice 3.1
1. 3x + 2y + z = 750
2. (a) Yes, because all three equations have three
variables, m, n and p where the power of the
variables is 1. The equation has zero n value.
(b) No, because there are equations with have the
variable of power of 2.
(c) Yes, because all three equations have three
variables, a, b and c where the power of the
variables is 1.
Self Practice 3.2
1. (a) x = 1, y = 3, z = 2
(b) x = −1, y = 2, z = 3
2. (a) x = −1, y = 3, z = −1
28
16
(b) x = – , y = 8, z =
3
3
t=3
Self Practice 3.3
t
0
3
(c) The graph of function h(t) with the value of
a = 2 is wider than the graph of r(t) with the value
of a = 4. Therefore, the bird that is represented
by the function r(t) moves at the highest position,
which is 36 m above the water level compared to
the bird that is represented by the function h(t)
with 18 m above the water level.
7. p = 3, q = 7
8. (a) b = −1 (b) c > 2 (c) c = 4
9. (a) 4 seconds
(b) 64 m
10. (a) (i) α
(ii) β
α+β
(iii) −αβ
(iv)
2
α+β
is the x-coordinate of the maximum point
(b)
2
of the graph and – αβ is the y-intercept of the
graph.
294
1. P = RM8 000, Q = RM2 000, R = RM14 500
2. Carnations = 80, roses = 50, daisies = 70
3. Pens = 3, pencils = 5, notebooks = 8
Intensive Practice 3.1
1. (a) x + y + z = 180, x – 20 = y + z, x – 10 = 3z;
100°, 50°, 30°
(b) x + y + z = 19, 2x + y + z = 22
x + 2y + z = 25; 3, 6, 10
2. (a) x = 2, y = 1, z = 0
(b) x = 3, y = 2, z = 1
8
44
(c) x = 5, y = −3, z = 1 (d) x = , y = – , z = –6
5
5
(e) x = −1, y = 3, z = 1 (f) No solution
3. Butters = 500, chocolates = 750,
coconuts = 900
4. Small = 9, medium = 6, large = 3
5. Chickens = 20, rabbits = 10, ducks = 20
Self Practice 3.4
2.
1. (a) x = 19, y = 31 and x = 2, y = −3
7
2
(b) x = – , y = and x = −4, y = 1
3
3
(c) x = 3.5811, y = −0.5811 and
x = 0.4189, y = 2.5811
14
(d) x = 7, y = −4 and x = – , y = 3
3
13
5
7
1
(e) x = , y = and x = , y =
4
8
2
2
(f) x = 3, y = 1 and x = −3, y = 7
y
2. (a)
3. h = −2, k =
(
1. (a) x + 2y + 3z = 120, 2x + 3y + 2z = 110
x + 4y + 2z = 180
(b) x + y + z = 30, 10x + 20y + 50z = 2 060,
x −3y – 2z = 25
2. (a) x = −2, y = 1, z = 3
(b) x = −1, y = 2, z = − 4
3. x = 15, y = 110, z = 55
1
2
4. h = 2; x = – , y =
7
7
5. RM 13 166.67, RM 6 666.67, RM166.66
6. 8 m, 15 m and 17 m
7. Yes, the line crossed the curve at another point,
5
9
which is x = , y =
2
2
8. 48 cm2
9. 4.5 m, 5.5 m
2
1
10. x = , y = 12 and x = , y = 24
3
3
7
28
11. Diameter = 7 m, radius = m; Diameter =
m,
2
9
14
radius =
m
9
10
5
(b)
2
3
4
5
x
(– 4.3, –1.7) and (4.0, 2.5)
y
6
5
4
3
2
Chapter 4 Indices, Surds and
Logarithms
1
–4
–3
–2 –1
0
–1
( )
)
MASTERY PRACTICE
20
15
1
1
; x = 1, y = −4
2
4. x = 5, y = 7
5. 35.8519 cm3 or 36 cm3
17 4
6. (−1, 0) and – ,
29 29
8 5
7. (−1, –2) and ,
3 3
30
25
–5 –4 –3 –2 –1 0
–5
–10
–15
–20
–25
( 65 , 35 ) and (3, −3)
1
2
3
–2
–3
–4
–5
(–1.2, –3.2) and (2.9, 0.9)
Self Practice 3.5
1. 8 cm, 9 cm
2. x = 8 cm, y = 6 cm or x = 6 cm, y = 4 cm
Intensive Practice 3.2
2
1. (a) x = 5, y = 3 and x = −6, y = –
3
(b) k = 3.7322, p = 1.5774 and
k = 0.2678, p = 0.4226
4
x
Self Practice 4.1
1. (a) 55x
(c) 9a(9–5 + 92)
(e) x y
6
11
(g) 27x y
2
(i) p7q20
(k)
2. (a)
(c)
5y
x10
2
6
1
6
a
1
a
17
20
(b)
(d)
(f)
(h)
(j)
(l)
1 1
–
75 73
7 8
cd
xy3
495
p10q3
x5y5
710
a4b2
6
18
(b) 4a 5
(d)
1 3 3
+ –
a a3 a4
Self Practice 4.2
1. (a) x = –11
(c) x = –3
2. (a) 10 cm
(b) x = –2
(b) 3.4868 cm
295
Intensive Practice 4.1
Self Practice 4.7
y3z2
xz3
(b)
x
y
(c) 4x12 y16
(d) e5 f 4
7y9
x3y5
(e) 5
(f)
x
78
x=7
x=4
m=1
1
2
3
→ 2 → 3 → −1 → −3 → −3
– → −2 → – →
2
3
10
(a) 5 904 900
(b) 5 minutes
79 570 057
RM51 874.85
1. (a)
2.
3.
4.
5.
6.
7.
8.
Self Practice 4.3
1. (a)
(c)
2. (a)
(b)
(c)
(d)
26
19
(b) 3
33
33
115
21
(d) 13
333
37
Surd because it is a non-recurring decimal.
Surd because it is a non-recurring decimal.
Not surd because it is a recurring decimal.
Surd because it is a non-recurring decimal.
Self Practice 4.4
6
1. (a) !w
(c) !w
9
8
(e)
3
(g) !w
4
!
(b) !w
15
(d) !w
30
(f) !w
6
(h) !w
10
Self Practice 4.5
260 = 2!w
65, (!w
16 !w
36)2 = 576,
1. !w
4!w
8
75
30!w
27
!w
= 2!w
= 5,
2,
= 15
4
3
2!w
3
6!w
!w
(!w
81)2 = 81
2. (a)
(c)
(e)
(g)
3
2!w
2!w
7
3!w
5
3!w
6
(b)
(d)
(f)
(h)
3!w
3
4!w
2
4!w
3
6!w
3
(b)
(d)
(f)
(h)
12!w
5
–12
3!w
7 – 35
20!w
7 – 154
Self Practice 4.6
1. (a)
(c)
(e)
(g)
(i)
2. (a)
(c)
(e)
296
5
8!w
2!w
7
4!w
5 + 25
87 + 35!w
3
–133
Not similar surds
Similar surds
Similar surds
(b) Similar surds
(d) Not similar surds
5
1. (a) 2!w
5
10
(c) !w
5
1
3
!w
+
(e)
6
2
16
+
3
w
!
(g)
23
2
(b) 7!w
2
1
(d)
4
15
+ 3!w
5 + 5!w
2 + !w
10
(f)
20
2 – 11!w
3
(h) 65 + 16!w
46
3 – 33!w
5 – 20!w
3
(i) 45!w
55
Self Practice 4.8
39 cm
1. !w
17
cm2
2. (a)
2
3. 13 + 4!w
3
4. (a) x = –2
(c)
1
4
(b) !w
66 cm
(b)
5
8
Intensive Practice 4.2
1. (a) !w
55
3
(c)
2
2. (a) 2!w
6
!
2
(c) 3!w
3. (a)
(c)
(e)
(g)
(i)
(k)
(m)
10
8!w
11!w
13
9!w
3 – 6!w
2
105!w
3
–!w
3
7!w
5 – 25
–154 – 20!w
7
(o) 4
(q)
4. (a)
(b)
(c)
(d)
5. (a)
(c)
(e)
6. (a)
(b) !w
70
(d) !w
6
(b) 9!w
2
4
(d)
4
!w
3
(b) 5!w
11
(d) 8!w
5
(f) 3!w
2 + 3!w
3
(h) 24!w
30
(j) 3!w
7 + 49
(l) 114 + 24!w
7
(n) 146 – 50!w
5
1
(p)
3
(r) 6
2
!w
5!w
5 + 7!w
3 – 7!w
7
3!w
5 + 5!w
3 + 18!w
2
13!w
5 + 21!w
3 – 14!w
7
11!w
5 + 17!w
3 – 7!w
7 + 36!w
2
2!w
5
(b) 3 + !w
5
5
(1 + !w
5)
2
!w
(d) !w
3 +
–
3
2
–5 + !w
21
17 + 7!w
5
(f)
2
4
3!w
7 −!w
2
–1
(b)
5
12 + !w
3
13
7. (2!w
5 – !w
2 ) cm
1 + 5!w
2
8. (a)
7
5. 3 years
6. 5.543 km
(c)
(b)
7 + 7!w
2
cm2
2
(b)
(d)
(b)
(d)
(b)
(d)
(f)
log2 128 = 7
log6 216 = 3
10−4 = 0.0001
43 = 64
1.9956
6
4
Self Practice 4.9
1. (a)
(c)
2. (a)
(c)
3. (a)
(c)
(e)
(g)
4. (a)
(c)
5. (a)
(c)
(e)
log3 81 = 4
log5 125 = 3
104 = 10 000
27 = 128
0.9542
–0.2375
4
5
x = 32
x = 256
138.7714
5568.0099
–0.0027064
(b) x = 512
(b) 24.6774
(d) 0.0004052
(f) 0.000027829
0.115
2.366
3
2
(b) 1.712
(d) –0.712
(b) 2
1. (a) log2 xy
2
(c) log2 xy3
(e) log3 m3n2
2. (a) 1 + q
1
(p + q)
(c)
2
()
x
(d) log ( 16!w
y )
x
(b) logb 3
y
4
3
(b) 2p + q
Self Practice 4.12
3. (a)
(c)
4. (a)
(c)
(
)
Self Practice 4.14
(ii) 2 amp
Intensive Practice 4.4
Self Practice 4.11
1. (a)
(c)
2. (a)
(c)
1. log51 = 0, log775 = 2.219
2. 2x + y – 3
3. 2
4
4.
3
6. 2p – m – 1
7. log2 2 + 1
x
8. y = 2x
1
9. (3 + x – y)
2
10. (a) 10−12 Watt
(b) 31 : 25
(c) 140 decibels
11. (a) 2 500 000
(b) 3 729 561
(c) Year of 2095
1. 3 weeks and 2 days
2. (a) 32 amp
(b) (i) 8 amp
(c) 3 seconds
Self Practice 4.10
1. (a)
(c)
2. (a)
(c)
Intensive Practice 4.3
2.8137
1.7959
2.7833
1.9820
2
t
2+t
t
2a + 3b
2
3+b
a+b
(b) 0.1550
(d) –0.1475
(b) 2.6309
3t
2
2 – 2t
(d)
t
a – 2b
(b)
3
(b)
(b) 2.1156 years
(b) 13219.2810 years
MASTERY PRACTICE
1. x = 0.4194
2. n = 2
35 + !w
21
!w
3.
2
4. t = 0
8
5.
2
12 + 8!w
6. (a) 59.05°C
7. 9 years
3 2
+
8.
2s t
5
9. x = , y = 2
2
10. 21.85 years
(b) 2.12 seconds
Chapter 5 Progressions
Self Practice 5.1
Self Practice 4.13
1. (a) 1.677
(c) 1.011
2. (a) 653803.075
(c) 1.982
(e) 1.792
3. 2 years
4. 11 years
1. (a) RM1 538.62
2. (a) 50 gram
3. (b) 6.93 hours
(b) 2.399
(b) −0.712
(d) 18.866
(f) 6.389, –8.389
1. (a)
(b)
(c)
(d)
2. (a)
(b)
(c)
(d)
14, add 14 to the previous terms.
3!w
3 , add 3!w
3 to the previous terms.
(p – q), add (p – q) to the previous terms.
loga23, add loga23 to the previous terms.
Arithmetic progression
Not an arithmetic progression
Not an arithmetic progression
Arithmetic progression
297
3. (a)
7. Company B, RM2 400
10
5
Self Practice 5.5
9
1. (a) Geometric progression
(b) Geometric progression
(c) Not a geometric progression
2. (a)
13
8
(b)
1
p
32
2
8
2p
–p
3p
4
7p
(b)
12p
(c)
17x
12x
7x
9x
x
11x
1. –23 → –12 → –3 → –0.4 → 25 → –14
2. (a) 37 years
(b) RM1 500
Self Practice 5.3
(b) 165
3
5
2. Horizontal
(a) 72
(b) 28 700
(c) 300
Vertical
(c) 31 570
(d) 22
(e) 30 100
3. 451 units
4. (a) 14 panels, the remaining wood pieces are 4
(b) White, 27 wood pieces
Self Practice 5.4
(b) 20 books
(b) 2 cm
Intensive Practice 5.1
An arithmetic progression
Not an arithmetic progression
12
(b) −9
12
(b) 14
425
(b) 4n[3n – 13]
1 225
−3
(b) 29
85
23
6. (a) −1
(b) –
2
298
1
6
1
6
1
12
1
12
1
24
1
12
1
24
1
48
4
1
625
243
→3→7→ →
→
→ 0.01
5
768
12
4
2. 23rd bounce
1.
Self Practice 5.2
1. (a)
(b)
2. (a)
3. (a)
4. (a)
(c)
5. (a)
(c)
1
3
Self Practice 5.6
4. Arithmetic progression.
1. (a) 21 days
2. (a) 2 cm
1
2
2
3. x = 3
The first three terms: 1, 4, 16 ; r = 4
5x
1. (a) 1 000
2
Self Practice 5.7
1. (a) 81.9
(c) 3 587 226.5
(b)
p[p22 – 1]
p2 – 1
2. 4
3. (b) 1 365
Self Practice 5.8
1. (a) 2 250
(c) 2 240
(b) RM30 000
(d) 53
Self Practice 5.9
1. (a) 400 cm
(b) 8 m
2. (a) The sequences of the perimeter: πj, πj(1.4),
πj(1.4)2, …
(b) 24.28 m
Intensive Practice 5.2
1. (a) n = 8, Sn = 1 640
6
(c) n = 9, Sn =
11
2. 9
1
3. (a) – , 3
2
1
4. (a) r = , a = 144 cm2
2
1
5. (a)
3
1
6. r = , T2 = 7.25 kg
2
(b) n = 7, Sn = log x−127
61
(d) n = 7, Sn = 5
64
(b) 512
(b) 71.72 cm2
(b) 351 cm
Self Practice 6.2
MASTERY PRACTICE
1. (a) 13
2. 3
3. (a) 4 cm3
1.
(b) −18
4. (a) a = 120, r =
(b) 324 cm3
1
2
y
30
(b) 240
5. (a) 56 chairs
(b) 572 chairs
6. (a) Savings of RM30 000 can be obtained.
(b) The savings does not reach RM30 000.
7. (a) 3
(b) RM5 460
25
20
15
10
5
Chapter 6 Linear Law
Self Practice 6.1
1. The graph of linear relation is Diagram 1(b). The
graph in Diagram 1(a) represents a non-linear
relation because the shape of the graph obtained is a
curve while the graph in Diagram 1(b) represents a
linear relation because a straight line is obtained.
2. (a)
Y
0
2.
4
10
15
20
25
30
x
L(cm)
Graph of L against m
3
2
0
45
20
40
60
80 100 120
m(g)
Self Practice 6.3
40
1. t =
35
2. (a)
30
29
1
x+
16
50
40
25
y
Graph of y against x
35
20
30
15
25
10
20
5
(b)
5
1
Graph of Y against X
50
Graph of y against x
15
0
2
4
6
8
10
12
X
10
5
Y
1.8
x
0
10 20 30 40 50 60
(b) y-intercept = 12.5, gradient = 0.375
(c) y = 0.375x + 12.5
Graph of Y against X
1.6
1.4
Self Practice 6.4
1.2
1. (a)
1.0
30
0.8
Graph of y against x
25
0.6
20
0.4
15
0.2
0
y
2
4
6
8
10
12
14
10
X
The graph (b) which is a straight line is a graph
of linear relation.
5
0
2
4
6
8
10
12
14
16
x
299
3. (a)
(b) (i)
(ii)
y-intercept = 4.0
y = 22
3
(iii) Gradient =
2
(iv) x = 7.4
3
(c) y = x + 4; y = 46
2
0.9
0.7
0.6
0.5
y
Graph of y
against x
0.4
0.3
8
0.2
6
0.1
4
(b)
(c)
4. (a)
2
–4
–2
0
–2
x
2
–4
(b)
y2
(i)
(ii)
(iii)
(i)
(ii)
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
m=1
0.15
1.512
4.9442
0.0619
xy
Graph of xy against x2
35
30
5
25
4
20
3
15
2
10
1
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
1–
x
The graph (a) is a non-linear graph while graph
(b) is a linear graph. The shape of graph (a) is a
curve while the shape of graph (b) is a straight
line.
Y
Graph of Y against X
5
0
(b) (i)
5
m=
(ii) 7.5
(iii) 10
(iv) 5.2
10
33
37
15
20
25
30
(c) x = 10.18
120
Self Practice 6.5
100
y
, X = 12 , m = −q, c = p
x2
x
y
(b) Y = , X = x, m = h, c = 1
x
(c) Y = yx2, X = x2, m = q, c = p
1. (a) Y =
80
60
40
20
0
Y=
300
log10 (x + 1)
40
Graph of y2 against 1–x
6
0
45
7
2.
Graph of log10 y against log10 (x + 1)
0.8
Intensive Practice 6.1
1. (a)
log10 y
10
20
11
518
X+
50
5
30
40
50
60
70
X
35
40
x2
2. (a)
(b)
1–
y
2.5
Graph of 1–y against
x
y
––2
x
y
Graph of ––2 against x
x
5.0
4.5
4.0
2
3.5
1.5
3.0
1
2.5
2.0
0.5
0
0.5
1
1.5
2
1.5
x
1.0
–0.5
0.5
–1
(b) (i)
0
q = −0.75
(iii) y =
(ii)
5
7
(a) y = 5x
+ 3x
2
y
=5
x2
3
+
x
x y!w
x=
(b) y = p!w
q
px + q
+
x
!w
Y-axis
X-axis
y
x2
1
x
3
1.2
1.0
5
0.8
0.6
y!w
x
x
p
q
0.4
0.2
log10a
(d) x = mxy x =
+ ny
y
mx + n
m
n
x
y
(e) ypx = q log10 y =
–log10 px log10 y
+ log10q
y
= ax + b
x2
Graph of log10 y against x
1.4
Gradient, Y-intercept,
m
c
b
2. (a)
x
2.0 2.5 3.0
1.6
(c) y = axb log10 y =
log10a + log10 y log10 x
b log10x
(f) y(b – x) x
=–x
= ax
y
a
b
+
a
1.5
log10 y
1.8
1.
Linear
1
(c) a = 1.504
b = 0.5
3. (a) log10y = b log10a + x log10a
(b)
31
p=
16
Intensive Practice 6.2
Nonlinear
0.5
x
y
x
0
–log10 p
log10q
2
(c) a = 2.0106
b = 0.5275
3
4
x
5
Self Practice 6.6
1. (a)
log10y
2.0
x
1
Graph of log10y against x
1.5
1.0
x
–1
a
b
a
0.5
0
2
4
6
8
(b) (i) p = 2.51189
(ii) q = 1.25893
(c) y = 7.94328
10
12
14
16
x
301
2. (a)
y
(c) N = 1.3416
(d) 20 workers
4. (a) log10L
Graph of y against xy
22
20
16
1
Graph of log10L against ––
14
T
18
16
14
12
10
12
8
10
6
4
2
8
6
4
2
xy
0 20 40 60 80 100 120 140 160
(b) a = –150.314868
b = –6.77094
(c) Gradient = 0.0066899, Y-intercept = 0.0452991
v
0.09
0.08
1. (a) p = 10
(b) p = 20
t2
2. (a)
0.07
0.06
0.05
Graph of t against p
2
0.04
200
0.03
180
0.02
160
0.01
140
120
1
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 ––
u
(b) (i)
100
v=
–10u
10 – u
10
y
4
=3+ 3
x
x
y
y
q
(b) 2 = px + q, 3 = p +
x
x
x
y p q
(c) xy = p + q x2, = 2 +
x x
p
p
(d) log10 y = log10 p + !w
x log10 k
(e) log10 y = log10 p + (x − 1) log10 k
(f) log10 y = x2 log10 k − log10 p
y
= px + q
(b) p = –0.25, q = 5.25
(a)
x
p = 100, q = 100
2
k = –1, h =
3
(a)
log2 y
1. (a) yx2 = 3x3 + 4,
60
40
20
0
10 20 30 40 50
(b) t = 15.5
(c) k = 2
3. (a) N2H
60
Graph of N2H against H
120
100
80
60
p
2.
3.
4.
5.
Qʹ(3, 5)
40
20
H
0
20 40 60 80 100
(b) a = 2.425
b = 12
302
(ii)
MASTERY PRACTICE
80
140
1
1
Graph of ––
v against ––
u
0.10
Intensive Practice 6.3
220
1
––
0 10 20 30 40 50 60 70 80 90 100 T
(b) (i) A = 1.585 × 108
(ii) b = 0.1258
(c) 221.7°C
5. (a)
1
––
Pʹ(1, 2)
x
0
(b) a = 1.414, b = 2.828
Pʹ(1, 2), Qʹ(3, 5)
11. (a)
8x + 3
x2
(b) y = 0.8850
6. (a) y =
7.
m
xy
16
Graph of xy against x
14
Graph of m against V
12
3
10
8
2
6
1
0
8. (a)
4
1
y
35
2
3
2
V
4
0
1
2
3
(b) (i) p = 4
(iii) y = 2.8571
(c) x = 0.0625
Graph of y against x
30
25
5
9. (a)
T°C
10 20
3
25
x+
8
2
30
40
50
60
x
Graph of T against t
40
30
q=6
Self Practice 7.2
1. (a) P(−3, 4)
(c) P(3, −1)
2. p = −2t
3. (a) C(4,4)
4. (a) 1 : 2, k = −2
(c) 1 : 4, k = 7
20
10
0
2
4
(b) 30.0
(c) (i) 28°C
(iii) 5 s
10. (a) log10 y
1.0
x
1. (a) Point P divides line segment AB in the ratio 1 : 2.
Point Q divides line segment AB in the ratio 1 : 1.
Point R divides line segment AB in the ratio
11 : 1.
(b)
S
B
A
2. (a) m = 2, n = 5
(b) P divides rope AB in the ratio 2 : 5.
(c) P(6, 0)
10
(b) y =
(ii)
6
Self Practice 7.1
15
0
5
CHAPTER 7 COORDINATE GEOMETRY
20
4
6
8
10
t(s)
(ii)
32.5°C
(b) P(−2, 1)
(b) D(2,1)
(b) 1 : 1, k = 5
(d) 2 : 3, k = 2
Self Practice 7.3
1. (28, 32)
2. (−1, 4), (2, 3)
3. (a) 2 : 1
Graph of log10 y against x
0.9
0.8
Intensive Practice 7.1
1. R(6, 4)
0.7
0.6
2. (a) Q(11, −2)
(b)
(152, 32 )
5. (a) U(5, −4)
(b)
(8,− 32 )
3. h = 7, k = 1
4. e = 10f
0.5
0.4
0.3
0.2
0.1
0
1
2
3
4
(b) (i) s = 1.90546
(iii) x = 4
(b) 5 units
5
(ii)
6
7
8
9
t = 1.20226
x
(c) 3 : 1
6. (a) 1 : 3
17
,4
7.
2
( )
(d) 5 units
(b) –1
303
1. (a) Parallel
(c) Perpendicular
1
2. (a) –
6
3. (a) 3
4. 8
(b) Parallel
(d) Perpendicular
(b) 2
(b) 6
Self Practice 7.5
1. 3y – 2x = 20
2. (a) (5, 5)
(b) 3.606 units
Intensive Practice 7.2
1. (a) Parallel
(b) Perpendicular
2. 3
3. (a) 3y + 2x = 23
(b) 2y – 3x = 11
S(1, 7)
4. (a) –9
(b) 17
5. h = −2
6. (a) 2y + x = 10, y = 2x (b) C(2, 4), 4.472 units
7. (a) AB is 3y – x = 5
DE is y + 3x = 15
(b) E(4,3), B(7,4)
8. (a) AB is parallel to CD, AB is perpendicular to AD,
CD is perpendicular to AD.
(b) 2y = x + 9
(c) y + 2x −22 = 0
9. (a) (i) 3y + 2x = 19
(ii) B(8, 1)
(b) (i) D(2, 5)
11. 2y + x = 17
(
(b) 12 units2
)
(c) 19 units2
2. k = −4
(b) 1, 21
(d) –7, 3
1
units2
2
1
(d) 27 units2
2
(b) 88
1
units2
2
2. 30 units2
1. 46
Self Practice 7.9
304
Self Practice 7.10
1. (a) x2 + y2 – 9 = 0
(b) x2 + y2 – 4x – 6y + 4 = 0
(c) x2 + y2 + 8x – 10y + 32 = 0
(d) x2 + y2 + 2x + 12y + 28 = 0
2. x2 + y2 + 4x − 2y − 20 = 0
3. (a) x2 + y2 + 8x = 0
(b) 4x2 + 4y2 + 29x + 5y + 26 = 0
(c) 5x2 + 5y2 + 36x − 56y + 164 = 0
(d) x2 + y2 – 10x + 4y + 21 = 0
4. 5x2 + 5y2 + 50x − 6y − 118 = 0
5. x2 + y2 + 12x = 0
6. 15x2 + 15y2 + 4x − 4 = 0
(b) 5x – 9y + 7 = 0
7. (a) x + 2y – 3 = 0
(c) 8x + 10y – 87 = 0
1. x2 + y2 − 6x − 8y + 9 = 0
(c) (7, 3), (12, 8)
2. (a) x − y – 4 = 0
1. (a) 3x2 + 3y2 + 12x − 68y + 364 = 0
26
, (0, 14)
(b) 0,
3
2
2
2. x + y −8x − 10y + 16 = 0
3. (a) x2 + y2 − 12x − 10y + 36 = 0
(b) 2, 10
4. y2 = 4x
5. (a) a2 + b 2 = 81
(b) 4x2 + y2 = 36
6.
1
2 –m
( )
P
2
A 1m D
1m
B
S
C
1– m
2
R
1 1–2 m
Self Practice 7.8
1. (a) C(7, 8), M(2, 4)
2. (a) k = 2
1
3. (a) 13 units2
2
1. (a) D(−2, 10), E(−1, 4) (b) 50 units2
2. (a) h = −2, k = −1
(b) 20 units2
3. (a) 0
(b) Point A, B and C are calinear.
1
4. 47 units2
2
5. 5, 37
6. 1, 5
7. (a) 20
(b) 14, 26
8. (a) k = 7
(b) (i) H(3, 11)
(ii) 1 : 2
9. (a) m = 2
(b) 17 units2
10. (a) 1.1402 km
(b) 0.645 km2
Intensive Practice 7.4
Self Practice 7.7
1. (a) 52 units2
Intensive Practice 7.3
Self Practice 7.11
Self Practice 7.6
1. (a) 24 units2
1
(c) 28 units2
2
5
2. (5, 0), – , 0
3
4. p = 1
11
5. (a) –3,
3
(c) –5, 7
(d) 27 units2
(c) E(7, −4)
Self Practice 7.4
(b) 1 : 4
(b) P(3,2)
(b) k = 1
Q
The locus consists of curves of a quadrant of 3 circles:
(i) APQ that is a quadrant with centre A and radius
1
of 2 m
2
(ii) BQR that is a quadrant with centre B and radius
1
of 1 m
2
(iii) CRS that is a quadrant with centre C and radius
1
of m
2
MASTERY PRACTICE
1. (a) h = −3, k = 5
(b)
2
5
(c) 2y + 5x = 16
2. (a) P(2, 2)
(b) y = x
1
4. (0, 6), (0, −3)
3. , 1
2
5. 2x2 + 2y2 + 19x + 35 = 0 6. (3, 3)
7. (a) C(4, −3)
(b) D(8, 7)
1
(c) (i) k = –
2
8. (a) P(3, 1)
(b) QR: y + 3x = 40 SR: 3y – x = 10
(c) Q(12, 4), S(5, 5)
(d) 25 units2
2
9. (a) 30 units
9k − 4h – 1
, 37 – 3k −2h (c) P(6, 5)
(b)
2
(d) y = x − 1
(e) (i) Q(8, 7)
(ii) 1 : 1
15
15 15
,
10. (a) R(−3, 6), S 0,
,T
2
8 4
9
(b) 18 units2
32
11. (a) h = 1, k = 4
(b) y + 2x = 10
(c) y = −2x + 8, y = −2x − 8
12. (a) y + 5x + 9 = 0
(b) P(–3, 6), D(7, 8), C(13, 4)
(c) 78 units2
13. (a) E(3, 1)
(b) Square
B(6, –3)
14. (a) P = 4x – 400
(b)
P
( ) (
)
2000
1600
1500
500
15.
(i)
y
100 200 300 350400 500
RM1 600
(ii)
x
350 copies
B(6, 7)
A (1, 2)
Self Practice 8.2
1. (a)
(b)
X
5N
S
4 cm
R
(c)
(d)
Y
1 cm represents 1 N
N
RS
1 cm represents 10 km
v
∼
4 cm
1 cm represents 5 km h–1
N
3.5 cm
a
1 cm represents 2 kg m s–1
2. !w
20 N, 026.57°
3. 117.15 km
→ → → → → →
4. MN = CD, EF = KL, GH = AB, a = d, c = f, b = e
→
→
5. (a) (i) ED
(ii) FE
→
(iii) AF
→
→
(b) (i) DC
(ii) CB
→
(iii) BA
Self Practice 8.3
1000
0
− 500
(c) Scalar quantity because the quantity only consists
of magnitude.
(d) Scalar quantity because the quantity only consists
of magnitude.
(e) Vector quantity because the quantity consists of
magnitude and direction.
P
C(7, 2)
0
x
Chapter 8 VECTORS
Self Practice 8.1
1. (a) Scalar quantity because the quantity only consists
of magnitude.
(b) Vector quantity because the quantity consists of
magnitude and direction.
→ 1
3
7 → 5
1. PQ = a~, ~x = – a~, y = – a~, RS = a~
2
2
4
4
Self Practice 8.4
→ 1→
1. AB = PQ
4
3. (a) m = –
3
,n=7
4
(b) m = 4, n = –3
→
7→
4. VW = XY
2
5. k = 4
→
8→
6. SR = – QT
5
Intensive Practice 8.1
→
1. AB = 3u
~
2. (a) 12 cm→
(b) (i) EC = 2a
~
1
1
4. h = – , k =
2
2
5. k = 4h – 2
(ii)
→
BE = 6b
~
305
(c) | p| = 5 units |q | = 8.602 units | r | = 5.099 units
~
~
~
Self Practice 8.8
Self Practice 8.5
1. (a)
(b)
2 u∼+ ∼v
(c)
(d)
u − 2 ∼v
∼
2. 131.19°, 106.30 km h−1
2
y
3. (a)
3~
1
(c)
y–x
3~ ~
4. (a) 578.27 km h–1
1. (a) 3.606 units
4
units
(c)
7
(e) 6 units
3i + 2j
2. (a) ~ ~
13
!w
1_ v + 2 u
∼
2∼
3v
u− _
2∼
2∼
(b) 8.062 units
(d) 13 units
(b)
(c) i
~
3. (a) Unit vector
(c) Unit vector
4. (a) ±1
(d)
30
7
→
→
2. (a) BD = –24x + 20y , AE = 6x + 15y
~
~
~
~
8.2
Intensive Practice
(e) ±0.866
5. p = ±3
(f)
1. (a) y + x
~ ~
1
(c) x + y
~ 2~
2. (a) 3x + y
~ ~
(c) –y + 2x
~
~
3. –a + b
~ ~ –1
5. 6.47 m s
1. (a)
(b) –x + y
~ ~
2
(d) – y – x
3~ ~
(b) 345.07°
(c) 0
Self Practice 8.6
1. k =
Self Practice 8.9
(c)
(b) y – 2x
~ ~
6. (a) (i)
–b + a
~ ~
2
3
(iii)
b+ a
5~ 5~
2
3
lb + la
(b) (i)
5 ~ 5 ~
5
2
(c) l = , µ =
6
3
3
(–b + a )
5 ~ ~
3
(iv) –b + a
~ 4~
(ii)
1.
3
(1 – µ)b + µa
~ 4 ~
306
( )
( )
(−9.52 )
( )
16
(–47
)
7
(d) (16)
(b)
(b) 2i + 6j
~ ~
(d) 5.5i + 20j
~
~
( )
30
10
2. Boat A = 15 , Boat B = 10
The two boat will not meet.
Intensive Practice 8.3
Self Practice 8.7
→
→
→
1. (a) OA = 2i + 2j , OF = –8i , BC = –10i + j ,
~ ~ →
~ ~
~ →
→
FA = 10i + 2j , DE = 14i , DO = –j
~ ~
~
~
→
2 →
–8 →
–10 →
10
(b) OA = 2 , OF = 0 , BC = 1 , FA = 2 ,
→
14 →
0
DE = 0 , DO = –1
→
5
2. (a) OB = 8
(b) 8.602 units
→
→
3. (a) (i) AB = 4i + j
(ii) BA = –4i – j
~ ~
→ ~ ~
→
(iii) BC = –i – 5j
(iv) DC = 2i
~ ~
~
→
→
(v) AC = 3i – 4j
(vi) AB = 4i + j
~ ~
→
→ ~ ~
(b) AB and DE, because they have the same gradient.
→
→
(c) BA and DE, because they are in the opposite
direction.
4. (a) p = 3i – 4j
(b) P(3, – 4), Q(–5, –7),
~ ~ ~
q = –5i – 7j R(1, 5)
~ ~
~
r = i + 5j
~ ~ ~
()
( )
()
(–930)
12
(–13
)
2. (a) 10i + 18j
~
~
(c) –7i – 26j
~
~
Self Practice 8.10
h = –10, k = 23
(ii)
(d)
6. h = ±!ww
2k w
−ww
k2
(b) –y + x
~ ~
4.
(b)
(d)
(b)
–i – 9j
~ ~
82
!w
–8i – 15j
~
~
17
Unit vector
Not a unit vector
±1
1
±
2
!w
±0.988
( )
( )
1. (a)
(38)
2. k = 10 or 1
23
3. m = , |u | : |v | = 9 : 16
3 ~ ~
→
4. (a) BC = 8i + 6j
~ ~
→
(c) AR = 6i + 2j
~ ~
5. 2.96 km h−1, 101.69°
(b) 8.544 N
(b)
4i + 3j
~ ~
5
6
6. (a) 4, −8
(b)
5
(c) 4
1
2
!w
7. ±
or ±
2
2
!w
5 i + !w
5j
8. –2!w
~
~
4–n
9. m =
4
10. (a) (50 – 4t)i + (20 + 4t)j
~
~
(b) After 5 hours
Intensive Practice 9.1
MASTERY PRACTICE
1. (a) a + b
(b) a – c
~ ~
~ ~
2
3. m = !ww
2. –
1 −wwnw2
3
2k + 17
4. h =
8
15i + 9j
~
~
(b) C(18, 13)
5. (a)
w
306
!ww
→
→ 2
7. BC = 2(u – v )
6. RS = (3i – 2j )
~ ~
5 ~ ~
8. (a) (i) –a + b
(ii) b – a
~ ~
~ ~
(iii) 2(b – a )
(iv) b – 2a
~ ~
~
~
(v) 2b – 3a
(vi) 2(b – 2a )
~
~
~
~
→ 1→
(c) Parallel
(b) AB = FC
2
20
9. (a) –21
(b) 29 km
30
(c) –32
10. (a) (i) 6u
(ii) 6u + 2v
~
~
~
(b) (i) 9u + 3v
(ii) 9u + (2 + 3k)v
~
~
~
~
1
k=
3
11. (a) (i) 4a + 4c
(ii) 3a + 3c
~
~
~
~
(iv) a + 3c
(iii) 4a + 6c
~
~
~
~
4
12. (a) The resultant velocity of Arul’s boat is 4i + j
~ 3~
7
The resultant velocity of Ben’s boat is 7i + j
~ 3~
The difference of the speed is 3.163 m s–1
3i – j
(b) ~ ~
10w
!ww
( )
( )
Chapter 9 Solution of triangles
1. ∠A = 64°, a = 37.359 cm, c = 26.158 cm
2. (a) BE = 8 cm, CE = 6 cm, DE = 15 cm
(b) ∠EAB = 53.13°, ∠BCE = 53.13°,
∠BCD = 126.87°, ∠ABD = 81.20°,
∠CBD = 25.06°
(c) Triangle BDC and triangle BDA have same
angle and two sides with same length.
3. (a) ∠PQR = 120°
(b) 5.529 cm
4. 61.62 cm
5. 138.58 m
Self Practice 9.5
1. (a) 3.576 cm
(c) 53.891 m
2. (a) 51.38°
(c) 99.06°
3. 69.93°
1. 29.614 m
3. 48.046 km
41.832 m
2.
4.
11.555 km
23.974 m
Intensive Practice 9.2
1. 4.071 cm, 6.475 cm
3. 46.50°
Self Practice 9.7
1. (a) 112.482 cm2
(c) 75.206 cm2
2. 27.078 cm
3. 51.237 cm2
4. 18.146 m2
(b) 28.670 cm2
Self Practice 9.8
1. 16.142 cm2
3. 2
1. 251.716 m2
p
q
r
=
=
sin P sin Q sin R
k
l
m
(b)
=
=
sin K sin L sin M
6
8
(c)
=
sin 40° sin 120°
2.
17.69 cm2
2.
66.169 cm2
Intensive Practice 9.3
Self Practice 9.2
(b) 52.29°
1.
2.
3.
4.
5.
6.
(a) 6 cm
43.012 cm2
7.501 cm or 17.713 cm
107.977 cm2
89.233 cm2
14.664 cm
(b) 6 cm2
Self Practice 9.10
Self Practice 9.3
Ambiguous case exists.
Ambiguous case does not exist.
57.86° or 122.14°
7.112 cm or 18.283 cm
Self Practice 9.4
1. 10.147 m
2.
Self Practice 9.9
1. (a)
1. (a)
(b)
2. (a)
(b)
(b) 35.26°
Self Practice 9.6
Self Practice 9.1
1. (a) 5.611 cm
(c) 9.331 cm
2. 55.344 m
(b) 18.661 cm
2.
41.224 m
2. (a) 19.519 cm
3. 98.13°, 3.5 units2
(b) 115.865 cm2
Intensive Practice 9.4
1.
2.
3.
4.
5.
(a) 40.20 cm2
9.266 km
(a) 31.241 cm2
31.455 km, 187.11°
457.803 m
(b) 125.63°
(b) Plane DBR
307
MASTERY PRACTICE
1. (a)
(b)
2. (a)
3. (a)
4. (a)
a = 6.504 cm, b = 5.239 cm
∠P = 105.03°, ∠Q = 49.92°, ∠R = 25.05°
6.756 cm
(b) 7.287 cm
13.82 cm
(b) 33.387 cm2
X
13.4 cm
10 cm
13.4 cm
(b)
5. (a)
6. (a)
7. (a)
(c)
Y
Y
42.2˚
10 cm
Z
Z1 X 42.2˚ 1
64.17°, 115.83°
5.903 cm
37.59°
118.9°
5.142 m
(c) 25.066 cm2
(b) 42.66°
(b) 14.309 cm2
(b)
B
A
C
C1
8. (a) 40°
(c) 5.763 cm2
9. (a)
N
(b) 3.875 cm
(b) 13.38 km
B
(c)
10. (a)
(b)
11. (a)
(b)
R
235˚
25
(i) 14.20 km
(i) 58.277 km
(iii) 2535.79 km2
Petrol station M
(i) 124.35°
(iii) 26.368 cm2
B
B
(ii)
(ii)
153.36°
58.76°
(c) 63.395 km
(ii) 6.943 cm
E
6.5 cm
C
9 cm
7 cm
Self Practice 10.2
1. 112
2. 104.76
Intensive Practice 10.1
1. I = 108.3
The average temperature in town P increased by
8.3% in February 2017 compared to January 2017.
2. I = 92.31
The price of a certain item decreased by 7.69% in the
year 2015 compared to the year 2012.
3. x = 0.5, y = 2.80, z = 125
4. p = 100, q = 131.90, r = 134.48
s = 125.86
5. 107.27
Self Practice 10.3
1. 105
2. 114
Self Practice 10.4
1. (a) IA = 150 , IB = 104, IC = 120, ID = 124
(b) 121
There was a 21% increase in the price of all
goods in the year 2016 compared to the year
2010.
(c) RM2.19
2. (a) a = 115, b = 150, c = 112.5, d = 33
(c) RM44.34
(b) 126.68
(d) 110
Intensive Practice 10.2
1. (a) 124
3. 76.4
4. (a) 130
(c) RM25.74
2.
93
(b) 132
MASTERY PRACTICE
D
A
10
(b) 17.762 cm2, 8.881 cm
33
(c) Triangle ZXY ʹ such that XZ is the same,
XY ʹ= XY, ∠XZY ʹ = ∠XZY
Triangle ZXY ʹ such that XZ is the same,
ZY ʹ = ZY, ∠ZXY ʹ = ∠ZXY
12. (a)
Chapter 10 INDEX NUMBERS
Self Practice 10.1
1. I = 82.20
The number of registered commercial vehicles
decreased by 17.80% in the year 2017 compared to
the year 2015.
2. I = 112.72
The average monthly expenditure of a household
increased by 12.72% in the year 2017 compared to
the year 2014.
308
3. 650 053 107 metric tonnes
4. 150
5. 94.48
1. (a) x = 1.00, y = 1.00, z = 110
(b) 112.5
2. m = 121, n = 122
3. (a) RM9.12
(b) 35 000
(c) 90.4%
(b) 0.935 million tonnes
4. (a) 64.12
(c) 87.15
5. (a) RM 15
(b) 187.5
(b) 105.25
6. (a) 4
7. (a) 133.03
8. p = 140, q = 130, r = 255
9. (a) 6.14 million
(b) 166.85
The number of visitors in the year 2020
increases by 66.85% compared to the year 2017.
10. x : y : z = 1 : 4 : 3
11. (a) P2014 = RM150.91, P2010 = RM188.64
(b) 12%
12. (a) 115
(b) RM198.38
Ambiguous case (Kes berambiguiti); Having
more than one solution in a triangle.
Base (Asas); If a is a number and is written in
index form, for example an, then a is the
base.
Base year (Tahun asas); A year that is chosen
as a starting point for the calculation of
a series of index numbers, usually a year
that has normal characteristics.
Codomain (Kodomain); A set, of which part
of it is mapped from a domain set.
Common difference (Beza sepunya); A
constant that is added to the previous term
to form an arithmetic progression.
Common ratio (Nisbah sepunya); A constant
that is multiplied to the previous term to
form a geometric sequence.
Composite function (Fungsi Gubahan);
Function of another function.
Conjecture (Konjektur); A prediction that
is not proven but seems true. If there is
sufficient proof, the prediction becomes a
theorem or a formula.
Continuous function (Fungsi selanjar); A
function with points on a graph that are
connected with a line or curve in a certain
interval.
Discrete function (Fungsi Diskret); A
function with points on a graph, which
points are not connected with a line or
curve.
Domain (Domain); Set of elements that are
mapped to another set by a relation.
Function (Fungsi); A special relation
in which every object in the domain
corresponds to exactly one element of the
range.
Heron formula (Rumus Heron); A formula
that is used to determine the area of a
triangle when the length of all sides are
known.
Horizontal line test (Ujian garis mengufuk);
A horizontal line that is used to identify if
a function is a one-to-one function.
Included angle (Sudut kandung); Angle
between two given sides of a shape.
Index (Indeks) If a is a number, n is a positive
integer and an, then n is the index.
Index number (Nombor indeks); A number
that expresses the relative change of a
quantity with respect to time.
Inverse function (Fungsi songsang); A
function that maps every image in the
function to its object.
Line of best fit (Garis lurus penyuaian
terbaik); The best straight line that is
drawn from points that do not form a
perfect straight line.
Line segment (Tembereng garis); A part of a
line that connects two end points.
Linear equation (Persamaan linear); An
equation that satisfies y = mx + c and
forms a straight line.
Linear equation in three variables
(Persamaan linear dalam tiga pemboleh
ubah); An equation in the form of
ax + by + cz = d, where a, b and c are
constants and non-zero.
Locus (Lokus); A point that moves with a path
that is traced by those points according to
certain conditions.
Logarithms (Logaritma); The logarithm of
a positive number N to a positive base
a is the index for a, i.e., if N = ax, then
loga N = x.
Nonlinear equation (Persamaan tak linear);
An equation whose highest power of a
variable is more than one.
One to one function (Fungsi satu dengan
satu); A relation whereby each object has
only one image.
309
Parallel line (Garis selari); Two or more
lines with the same gradient.
Parallel vector (Vektor selari); Two vectors
are parallel if one vector is a scalar
multiple of another vector.
Perpendicular line (Garis serenjang); Two
lines that intersect at 90 degrees.
Plane (Satah); A flat surface that consists
of a horizontal plane, vertical plane and
curved plane.
Power (Kuasa); If a is a number and n is a
positive integer, then an is a number and is
called a power of n.
Price index (Indeks harga); A statistical
measurement that is used to show price
changes in a certain time.
Progression (Janjang); A sequence of
numbers that is formed by adding or
multiplying a constant to the previous
term (except the first term).
Quadratic function (Fungsi kuadratik); A
function in the form of f(x) = ax2 + bx + c,
where a, b and c are constants and a ≠ 0.
Quadratic inequality (Ketaksamaan
kuadratik); An inequality with one
quadratic expression with a variable on
the left and zero on the other side.
Quadratic equation (Persamaan kuadratik);
An equation in the form of ax2 + bx + c = 0,
where a, b and c are constants and a ≠ 0.
Resultant vector (Vektor paduan); A single
vector that is formed from a combination
of two or more vectors.
Range (Julat); Subset of the codomain that
contains all the images which have been
310
mapped by the objects in the domain.
Root (Punca); The value of an unknown
which satisfies an equation.
Simultanenous equation (Persamaan
serentak); Two or more equations that
contain common variables.
Tangent (Tangen); A straight line that touches
a curve at a point without crossing over.
Term (Sebutan); Numbers that form a
numerical sequence or progression.
Terminal point (Titik terminal); The endpoint
on a line segment that represents a vector.
Unit vector (Vektor unit); A vector with
magnitude of one unit at a certain
direction.
Variable (Pemboleh ubah); A quantity whose
value is unknown and not fixed.
Vector magnitude (Magnitud vektor); The
length or size of a vector.
Vector quantity (Kuantiti vektor); A quantity
that has magnitude and direction.
Vertex (Verteks); The minimum or maximum
point of a parabola.
Vertical line test (Ujian garis mencancang);
A vertical line that is used to determine
whether the relation of a graph is a
function.
Weightage (Pemberat); A constant assigned
to a single item, indicative of the item’s
relative importance.
Zero vector (Vektor sifar); A vector that has
zero magnitude and undefined direction.
Chow W. K. (2013). Discovering Mathematics (2nd ed.). Singapore: Star Publishing Pte Ltd.
Greenwood, D., Robertson, D., Woolley, S., Goodman, J. & Vaughan, J. (2017). Essential
Mathematics for the Australian Curriculum Year 10. Australia: Cambridge University Press.
Thomas, E. J. & Brunsting, J. R. (2010). Styles and Strategies for Teaching Middle School
Mathematics. USA: Corwin Press.
Ho, S. T. & Khor, N. H. (2001). New Additional Mathematics. Singapore: SNP Panpac Pte Ltd.
Istilah Matematik untuk Sekolah-sekolah Malaysia (2003). Kuala Lumpur, Malaysia: Dewan
Bahasa dan Pustaka.
Yeo, J., Keng, S. T., Cheng, Y. L & Chow, I. (2013). New Syllabus Additional Mathematics
(9th ed.). Singapore: Shinglee Pte Ltd.
Rondie, P. L., Kemp, E., Buchanan, L., Fensom, J. & Stevens, J. (2012). Oxford IB Diploma
Programme: Mathematics Standard Level Course Companion. UK: Oxford University Press.
Lim, L. N. (2007). GCE O Level Additional Mathematics Key Points Exam Guide. Singapore:
Redpost Publications Pte Ltd.
Sullivan, M. (1996). Algebra & Trigonometry (4th ed.). USA: Prentice Hall, Inc.
Allen, R. G. D. (1975). Index Numbers in Economic Theory and Practice. USA: Transaction
Publishers.
O’Neill, R., Ralph, J. & Smith, P. A. (2017). Inflation: History and Measurement. UK: Palgrave
Macmillan.
Barret, R. (2008). NCEA Level 2 Mathematics Year 12. New Zealand: ESA Publications (NZ) Ltd.
Pemberton, S. (2016). Cambridge IGSCE and O Level Additional Mathematics Coursebook.
UK: Cambridge University Press.
Afriat, S. N. (2014). The Index Number Problem: Construction Theorems. UK: Oxford University
Press.
Zaini Musa, Abdul Rahim Mohd Idris & Tee, H. T. (2011). Matematik Tambahan Tingkatan 4.
Shah Alam: Cerdik Publications Sdn. Bhd.
311
Algebra
13, 27, 90, 115
Arithmetic progressions 128, 129, 130, 133,
134
Axis of symmetry
50, 54, 56, 57, 58
Cartesian plane 178, 182, 184, 190, 192,
204, 207
Common difference 128, 129, 130, 133,
134, 135
Line segment 176, 181, 183, 206
Linear equation 70, 72, 73, 75, 78
Magnitude 212, 214, 224, 227, 230
Mapping
Mass
2, 13, 21, 22, 31
2, 123, 213, 218
Minimum point 49, 50, 54, 58, 63
Notation 2, 3, 14, 31
Composite 15, 17, 23, 30, 279, 280, 283,
284, 287
Object
Constant 37, 46, 54, 56, 57, 63, 64, 128, 140
Plane 71, 72, 73
Cosine rule
252, 254, 263
Discriminant 45, 46, 51, 52, 60, 65
Elimination method 73, 75, 78, 80, 81, 85
Factorisation method 80
Formula 37, 45, 55, 67, 94, 106, 109, 120,
122
Function 3, 4, 9, 12, 20, 24, 27, 109, 111
Gradient 154, 157, 158, 160, 163, 166, 184,
186, 188, 206, 209
Graph
Image
43, 49, 50, 109, 111
3, 9, 15, 30
Index 90, 109, 110, 274, 276, 277, 279
Intersection point 36, 37, 57, 60, 72, 82, 87
Inverse
20, 24, 27, 30
Laws of logarithms 119, 109, 113
Line of best fit 156, 157, 159, 160, 164, 166,
168, 170, 173
312
3, 9, 15, 30
Perpendicular line 184, 206
Pythagoras Theorem 242, 252
Quadratic 36, 38, 41, 45
Relation 2, 5, 20, 117, 125
Scalar quantity 212, 213, 214
Sequence 128, 129, 130, 135, 140, 143
Sine rule
Squares
242, 244, 252, 263, 266
37, 45, 55
Statistics 272, 277, 279
Straight line 154, 156, 159, 160, 162, 170
Substitution method 73, 76, 80, 85
Variable 2, 27, 28, 70, 72, 79, 80
Vector 212, 213, 214, 215, 217, 218, 221,
225, 227
Velocity
212, 213, 215, 223, 234
Vertex 54, 55, 57, 60
Weightage 279, 280, 284, 287
Dengan ini SAYA BERJANJI akan menjaga buku ini
dengan baiknya dan bertanggungjawab atas kehilangannya,
serta mengembalikannya kepada pihak sekolah pada
tarikh yang ditetapkan
Skim Pinjaman Buku Teks
Sekolah
Tahun
Tingkatan
Nama Penerima
Nombor Perolehan:
Tarikh Penerimaan:
BUKU INI TIDAK BOLEH DIJUAL
Tarikh
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RM9.90
ISBN 978-967-2375-42-5
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